
Class __ilS£>Q5* 
Book_J±^2i . 

SMITHSONIAN DEPOSIT 



/ 



THE 



CARPENTER'S NEW GUIDE: 



■Strata 




BEING A 



COMPLETE BOOK OF LINES 



FOR 



CARPENTRY AND JOINERY. 



TREATING FULLY ON 



SOFFITS, BRICK AND PLASTER GROINS, NICHES OF EVERY DESCRIPTION, 
SKY-LIGHTS, LINES FOR ROOFS AND DOMES; 



WITH A GREAT VARIETY OF 



DESIGNS FOR ROOFS, TRUSSED GIRDERS, FLOORS, DOMES, BRIDGES, &c; 
ANGLE BARS FOR SHOP FRONTS, &c; AND RAKING MOULDINGS. 

ALSO 

ADDITIONAL PLANS FOR VARIOUS STAIR-CASES, WITH THE LINES FOR PRODUCING THE FACE AND 

FALLING MOULDS NEVER BEFORE PUBLISHED, AND GREATLY SUPERIOR TO THOSE 

GIVEN IN THE FORMER EDITION OF THIS WORK, 

BY WILLIAM JOHNSTON, ARCHITECT, 

OF PHILADELPHIA. 

THE WHOLE FOUNDED ON TRUE GEOMETRICAL PRINCIPLES; 

THE THEORY AND PRACTICE WELL EXPLAINED AND FULLY EXEMPLIFIED ON 

EIGHTY-THREE COPPER-PLATES: 

INCLUDING 

SOME OBSERVATIONS AND CALCULATIONS ON THE STRENGTH OF TIMBER. 



7 



BY PETER NICHOLSON, 

AUTHOR OF "THE CARPENTER AND JOINER'S ASSISTANT," " THE STUDENT'S INSTRUCTOR TO THE FIVE ORDERS,'' ETC. 

THIRTEENTH EDITION. 

PHILADELPHIA: 
GRIGG, ELLIOT AND CO., No. 14 NORTH FOURTH STREET. 

1848. 






"1/. 






Entered according to the Act of Congress, in the year 1848, by 

GRIGG, ELLIOT AND CO., 

in the Clerk's Office of the District Court for the Eastern District of Pennsylvania. 




PHILADELPHIA : 
T. E. AND P. G. COLLINS, PRINTERS. 



PREFACE. 



To a book intended merely for the use of Practical Mechanics, much Preface is not 
necessary : — It is proper, however, to say, that whatever rules by previous authors have on 
examination proved to be true and well explained, these have been selected and adopted, 
with such alterations as a very close attention has warranted for the more easily compre- 
hending them, for their greater accuracy or facility of application ; added to these, are many 
examples which are entirely of my own invention, and such as will, I am persuaded, con- 
duce very much to the accuracy of the work and to the ease of the workman. 

The arrangement of the subjects in this work is gradual and regular, and such as a stu- 
dent should pursue who wishes to attain a thorough knowledge of his profession : and as it 
is Geometry that lays down all the first principles of building, measures of lines, angles, and 
solids, and gives rules for describing the various kinds of figures used in buildings ; there- 
fore, as a necessary introduction to the art treated of, I have first laid down, and explained 
in the terms of workmen, such problems of Geometry as are absolutely requisite to the 
well understanding and putting in practice the necessary lines for Carpentry. These pro- 
blems duly considered, and their results well understood, the learner may proceed to the 
theoretical part of the subject, in which Soffits claim particular attention ; for, by a thorough 
knowledge of these, the student will be enabled to lay down arches which shall stand exactly 
perpendicular over their plan, whatever form the plan may be : on this depends the well 
executing all groins, arches, niches, &c, constructed in circular walls, or which stand upon 
irregular bases ; wherefore the importance of rightly understanding these I cannot too much 
insist upon, their construction being so various and intricate, and their uses so frequently 
required. The two plates of cuneoidal or winding soffits are new, and are constructed in a 



iv PREFACE. 

more simple and more accurate manner : yet this method is only a nearer approximation to 
truth than the former one ; the surface of a conchoid cannot be developed ; that is, it cannot 
be extended on a plane : it is therefore absurd to look for perfection on this subject. 

The next subject which regularly presents itself is Groins ; for the construction of 
which there will be found many methods entirely new ; and besides the common figures, I 
have shown many which are difficult of execution, and not to be found in any other author. 
I have displayed a variety of methods for constructing spherical niches, a form more fre- 
quently wanted than the elliptic, which only has yet been explained. 

Among the various methods for finding the Lines for Roofs, I have given an entire new 
one for finding the down and side bevels of puriines, so that they shall exactly fit against the 
hip rafter; and by the same method the jack rafter will be made to fit. 

Of Domes and Polygons, I have shown an entirely new method for finding their cover- 
ing, within the space of the board, thereby avoiding the tedious and incommodious method 
of finding the lines on the dome itself, as has been always practised heretofore : also a method 
for finding the form of the boards near the bottom, when a dome is to be covered horizon- 
tally. Of dome-lights over stair-cases, or in the centre of groins, a rule upon true principles 
is given, for finding their proper curve against the wall, and the curve of the ribs ; this has 
never before been made public. 

Having gone thus far in the Art of Carpentry, it is necessary for me to say, by way of 
caution and guard to the ardent theorist, that there are some surfaces which cannot be de- 
veloped ; such as spherical or superoidical domes, where their coverings cannot be found by 
any other means than by supposing the curved surface to become polygonal ; in which case 
such domes may be covered upon true principles, as may be demonstrated. Let us suppose 
a polygonal dome inscribed in a spherical one ; then, the greater the number of sides of the 
polygonal dome, the nearer it will coincide with its circumscribing spherical one. Again, 
let us suppose that this polygonal dome has an infinite number of sides ; then, its surface 
will exactly coincide with the spherical dome, and therefore in anything which we shall have 
occasion to practice, this method will be sufficiently near ; as, for example, in a dome of one 
hundred sides, of a foot each, the rule for finding such a covering will give the practice so 
very near, that the variation from absolute truth could not be perceived. 



PREFACE. v 

Having gone through the constructive part of Carpentry, I next proceed to examples 
showing the best forms of floors, partitions, trusses for roofs, truss girders, domes, &c, which 
shall resist their own weight, or the addition of any adventitious load. 

To conclude : as I pretend not to infallibility, I hope to be judged with candor, being 
always open to conviction, from a knowledge of the difficulty and intricacy of science ; yet 
I hope that my labors may be of some use to others in shortening the road, and smoothing the 
path through which, for many years, I have been a persevering traveler for knowledge : I 
shall then be satisfied, and not deem my time misspent if my labors tend to the public good. 

P. NICHOLSON. 



CONTENTS. 



PRACTICAL GEOMETRY. 

PAGE 

Definitions -.-----.---10 

Problems ----------- 12 

CARPENTRY. 

Linings of Soffits - - - - - - - -- - 23 

Kirb Lights for Church Work -------- 28 

Groins - - - - - - - - - -29 

Niches ----------- 39 

Angle Brackets for Plaster Cornices -,- - - - - - - 43 

Pendentives ---------- 44 

Roofing, Plain and Spherical - -- - - - . - - - 45 

Strength of Timber --------- 51 

Designs for Roofs - - -•- - - - - - - 64 

JOINERY. 

■ 

Hand-railing and Stair-casing -------- 75 

Diminishing Columns - - - - - - - - . - - 100 

Cylindro-cylindric Sash Work, or Circular-headed Sash Work, in Circular Walls - - 101 

Angle Bars for Shop Fronts --------- 103 

Raking Mouldings ---------- 104 

The method of proportioning Cornices - - - - - - - - 105 

Mouldings upon the Spring -------- 106 

Sky-lights - - - - - - - - - - - 107 

CONCLUSION. 

The Ellipsis ---------- 109 

Raking Mouldings - - - - - - - - - -110 

Diminishing Columns --------- m 



PRACTICAL GEOMETRY. 



GEOMETRY is the science of extension and magnitude; it teaches the construction 
of all right-lined and curvilineal figures, and is divided into Theoretical and Practical. 

The Theoretical part is founded upon reason and self-evidence : it demonstrates the 
construction of variously formed figures, and evinces the truth, or detects the falsehood on 
which they are made. This is the foundation of the Practical part ; and without a know- 
ledge of the Theory, no invention to any degree certain can be made in the Practice. 

The uses of Geometry are not confined to Carpentry and Architecture, but in the 
various branches of the Mathematics, it opens and discovers to us their secrets. It teaches 
us to contemplate truths, to trace the chain of them, subtile and almost imperceptible as it 
frequently is, and to follow them to the utmost extent. 

Its uses are great and necessary in Astronomy and Geography. The science of Per- 
spective is entirely dependent upon its principles. To enumerate its many uses is beyond 
my power. Those who desire to become thoroughly acquainted with Geometry, will do 
well to study attentively the elements of Euclid. 

As my labors are not intended for the abstruse Mathematician, but for the instruction of 
the Practical Carpenter, I shall omit all speculative demonstrations, the sections of Cylinders 
and Globes excepted (which are not to be found in Euclid), and confine myself to the useful 
part of the science, viz. Practical Geometry. 
2 



10 PRACTICAL GEOMETRY. 



PLATE 1. 

DEFINITIONS. 

1. A POINT has position but not magnitude. 

2. A line is length without breadth or thickness. 

3. A superficies hath length and breadth only. 

4. A solid is a figure of three dimensions , having length, breadth, and thickness. 

Hence surfaces are the extremities of solids, and lines the extremities of surfaces, and points the extremities 
of lines. 

5. Lines are either right, curved, or mixed of these two. 

6. A right or straight line lies all in the same direction between its extremities, and is the shortest distance 
between two points, as A. 

7. A curve continually changes its directions between the extreme points, as C. 

8. Lines are either parallel, oblique, perpendicular, or tangential. 

9. Parallel lines are always at the same distance, and will never meet, though ever so far produced, as D 
and E. 

10. Oblique right lines change their distance, and would meet, if produced, as F. 

11. One line is perpendicular to another when it inclines no more to one side than another, as G. 

12. A straight line is a tangent to a curve when it is produced and touches it without cutting, as H. 

1 3. An angle is the inclination of two lines towards one another in the same plane, meeting in a point, as I. 

14. Angles are either right, acute, or oblique, as K. 

lb. A right angle is that which is made by one line perpendicular to another, or when the angles on each 
side are equal, as G. 

16. An acute angle is less than a right angle, as K. 

17. An obtuse angle is greater than a right angle, as L. 

18. A superficies is either plane or curved. 

19. A plane, or plane surface is that to which a right line will every way coincide ; — but if not, it is 
curved. 

20. Plane figures are bounded either by right lines or curves. 

21. A solid is said to be cut by a plane when it is cut through in any particular place, and the place that 
is cut is called the section of the solid. 

22. Plane figures, bounded by right lines, have names according to the number of their sides, or of their 
angles, for they have as many sides as angles — the least number is three. 

23. An equilateral triangle is that whose three sides are equal, as M. 

24. An isosceles triangle has only two sides equal, as N. 

25. A scalene triangle has all sides unequal, as O. 

26. A right-angled triangle has one right angle, as P. 

27. Other triangles are oblique' angled, and are either obtuse or acute. 

28. An acute-angled triangle has all its angles acute, as M or N. 

29. An obtuse-angled triangle has one obtuse angle, as O. 

30. A figure of four sides and angles is called a quadrangle, or quadrilateral, as Q, R, S, T, U, and V. 



Flate 1. 








C 3 




! B 




PRACTICAL GEOMETRY. 11 

31. A parallelogram is a quadrilateral, which has both pairs of its opposite sides parallel, as Q, R, U, and 
V; and takes the following particular names. 

32. A rectangle is a parallelogram having all its angles right ones, as Q and R. 

33. A square is an equilateral rectangle, having all its sides equal, and all its angles right ones, as Q. 

34. A rhombus is an equilateral parallelogram, whose angles are oblique, as U. 

35. A rhomboid is an oblique-angled parallelogram, as V. 

36. A trapezium is a quadrilateral which has neither pair of its sides parallel, as T. 

37. A trapezoid hath only one pair of its opposite sides parallel, as S. 

38. Plane figures having more than four sides are in general called polygons, and receive other particu- 
lar names according to the number of their sides or angles. 

39. A pentagon is a polygon of five sides, a hexagon has six sides, a heptagon seven, an octagon eight, a 
nonagon nine, a decagon ten, an undecagon eleven, and a dadecagon twelve sides. 

40. A regular polygon has all its sides and its angles equal ; and if they are not equal, the polygon is 
irregular. 

41. An equilateral triangle is also a regular figure of three sides, and a square is one of four ; the former 
being called a trigon, and the latter a tetragon. 

42. A circle is a plane figure bounded by a curve line called the circumference, which is everywhere equi- 
distant from a certain point within, called its centre. 

43. The radius of a circle is a right line drawn from the centre to the circumference, as a b at W. 

44. A diameter of a circle is a right line drawn through the centre, terminating on both sides of the cir- 
cumference, as c d at W. 

45. An arch of a circle is any part of the circumference. 

46. A chord is a right line joining the extremities of an arch, as a b at X. 

47. A segment is any part of a circle bounded by an arch and its cord, as X. 

48. A semicircle is half the circle, or a segment cut off by the diameter, as Y. 

49. A sector is any part of a circle bounded by an arch and two radii, drawn to its extremities, as Z. 

50. A quadrant, or quarter of a circle, is a sector having a quarter of the circumference for its arch, and 
the two radii are perpendicular to each other, as A 1. 

51. The height or altitude of any figure is a perpendicular let fall from an angle, or its vertex, to the 
opposite side, called the base, as a b at B 2. 

52. When an angle is denoted by three letters, the middle one is the place of the angle, and the other two 
denote the sides containing that angle ; thus, let a b c be the angle at C 3, b is the angular point, and a b, and 
b c are the two sides containing that angle. 

53. The measure of any right-lined angle is an arch of any circle contained between the two lines which 
form the angle, and the angular point being in the centre, as D 4. 



12 PRACTICAL GEOMETRY. 



PLATE 2. 



PROBLEMS. 



Figure 1. To draw a Perpendicular to a given Point in a Line. 

A B is a line, and c a given point ; take a and b, two equal distances on each side of c, 
and with the foot of the compasses in a and b make an intersection d, and draw d c, which 
is the perpendicular. 

Fig. 2. To make a Perpendicular with a Ten Foot Rod. 

Let a b be six feet, then take eight feet, and from a make an arch at c b, and from the 
point a with the distance of ten feet across at c, then draw c b, which is the perpendicular. 

Fig. 3. To let fall a Perpendicular from a given Point to a Line. 

In the given point c make an arch to cross the line in a and b, and from a and b make 
an intersection at d, and draw c d the perpendicular. 

Fig. 4. To draw a Perpendicular upon the End of a Line. 

Take any point d at pleasure above the line, and with the distance d b make an arch a 
b c, and draw a line a d to cut it at c, and draw c b the perpendicular. 

Fig. 5. To divide a Line into two equal Parts by a Perpendicular. 

From the extreme points a and b describe two arches to intersect at c and d, draw c d, 
which divides the line into two equal parts. 

Fig. 6. To divide any given Angle into two equal Angles. 

Take two equal distances a b and a c on each side of the angular point a, and with the 
same opening of the compasses or any other, place the foot in b and c, make an intersection 
at d, and draw d a, which will divide the angle into two equal parts. 

Figs. 7 and 8. An Angle being given, to make another equal to it, from a given Point 

in a right Line. 

Let b a c be the angle given, and c d a right line, c the given point, on a make an arch 
b c with any radius, and on c with the same radius describe an arch d e, take the chord of 
b c, set it from d to e, and draw e c, then the angle e c d will be equal to c a b. 



,L 



Fiq:l. 



Fla(,> 2. 



Fig. 2. / 



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Fid. 3. 



Fig. 4. / 

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Plate 3. 




b a 









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h 





PRACTICAL GEOMETRY. 13 



PLATE 3. 

Fig. 1. Upon a right line to make an equilateral Triangle. 
Take a b the given side from a and b, and make an intersection at c, and draw c a and c b. 

Fig. 2. Upon a right Line to make a Square. 

With the given side a b, and in the points a and b, describe two arches to intersect at e, divide b e into 
two equal parts aty, make e d and e c each equal to ef, draw a d, d c, and c b. 

Figs. 4, 5, 6. T/;e side of any Polygon being given, to describe the Polygon to any Kumber of sides whatever. 

On one extreme of the given side make a semicircle of any radius, but it will be most convenient to make 
it equal to the side of the polygon; then divide the semicircle into the same number of equal parts as you 
would have sides in the polygon, and draw lines from the centre through the divisions in the semicircle, 
always omitting the two last, and run the given side round each way upon these lines, join each side, and 
it will be completed. 

Example in a Pentagon. Fig. 4. 

Let a b be the given side, and continue it out to c ; on a, as the centre with the radius a b, describe a semi- 
circle, divide it into five equal parts; through 2, 3, 4, draw a 2, a d, a e ; make b e equal to a 6, 2 d equal 
to 2 a or a b; join 2 d, d e, and e b. In the same manner may any other polygon be described. 

N. B. This depends upon the equality of the angles upon equal arcs. See Fig. 3. 

Fig. 7. Through a given Point a, to draw a Tangent to a given Circle. 
Draw a o to the centre, then through a draw b c perpendicular to a o, it will be the tangent. 

Fig. 8. A tangent Line being given, to find the Point where it touches the Circle. 

From any point a in the tangent line b a, draw a line to the centre o, and divide a o into two equal parts 
at m and with a radius m a, or m o, describe an arch, cutting the given circle in n, which is the point re- 
quired. 

Fig. 9. Two right Lines being given, to find a mean Proportion. 

Join a b and b c in one straight line, divide a c into two equal parts at the point o, with the radius o a or o c 
describe a semicircle, and erect the perpendicular b d, then is a b to b d as b d is to b c. 

Fig. 10. Through any three Points to describe the Circumference of a Circle. 

From the middle point b draw chords b a and b c to the two other points a and c, divide the chords a b 
and b c into two equal parts by perpendiculars meeting at 0, which will be the centre. 

To find the length of any Arc A B C of a Circle. 

Draw the chord A C and produce it to E ; bisect the arc A B C in B, and make A D equal to twice A B ; 
divide CD into three equal parts, and set one out to E; then A E is the length of the arc. 



14 PRACTICAL GEOMETRY. 



PLATE 4. 

Fig. 1. Three Lines being given, to form a Triangle. 

Take one of the given sides a b, and make it the base of the triangle ; take the other 
side a c, and from a, describe an arch at c ; then take the third side b c, and from b, describe 
another arch crossing the former at c, and join a c and b c. 

Note : that any two lines must be greater than a third. 

Figs. 2, 3. To make a Quadrangle equal to a given Quadrangle. 

Divide the given quadrangle,^. 2, in two triangles; make the triangle efg equal to 
a b c, and e g h equal to a c d, and it is done. 

Figs. 4, 5. Any irregular Polygon being given, to make another of the same dimensions. 

Divide the given polygon,^. 4, into triangles, and mfg. 5, make triangles in the same 
position, respectively equal to those infg. 4; then will the irregular polygon f g, h, i, k, be 
equal and similar to a b c d e. 

Fig. 6. To make a Rectangle equal to a given Triangle. 

Draw a perpendicular c d, divide it into two equal parts at e, through e dr&wfg, parallel 
to the base a b, draw a fb g, perpendicular; then will the rectangle a b gf be equal to 
the triangle a b c. 

Fig. 7. To make a Square equal to a given Rectangle. 

Let a b c dbe the given rectangle; continue one of its sides as a b out to e, make b e 
equal to the other side b c, divide a e into two equal parts at i, with the radius i e or i a 
make a semicircle afe, and draw bf perpendicular to a b; make the square bfg h, which 
is equal to the parallelogram abed. 

Fig. 8. To make a Square equal two to given Squares. 

Make the perpendicular sides a c and a b of the right-angled triangle cab equal to the 
sides of the given squares A and B, draw the hypothenuse c b, which is the side of the 
square D equal to the squares A, B, C. 



Plate 4. 




Fig. 4. 










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PRACTICAL GEOMETRY. 15 



PLATE 5. 

Fig. 1. To draw a Segment of a Circle to any Length and Height. 

a b is the length, i h the height; divide the length a b into two parts by a perpendicular 
g c; divide h by the same method, then their meeting at g will be the centre ; fix the foot of 
the compasses in g, extend the other leg to h, make the arch a h b, which is the segment. 

Fig. 2. To draw a Segment by Rods, to any Length and Height. 

Make two rods c e and cf to form an angle e c f so that each may be equal to a b, the 
opening ; place the angle c to the height, and the edges to a and b, put a piece a b across 
them to keep them tight, then move your lath round the points a, b, and it will describe the 
segment at the point c. 

Fig. 3. To describe a Segment of a Circle at twice, upon true Principles, by aflat Triangle. 

Let the extent of the segment be a b, its height c d, from the extreme b to the top d 
draw b d, through the point d draw e d parallel to the base a b, equal in length to d b, de- 
scribe one half, as you see at G ; then move your nail, or pin, out of a, stick it in the 
point b, and describe the other half. 

Fig. 4. The transverse and conjugate Axis of an Ellipsis being given, to draw its Re- 
presentation. 

Draw a d parallel and equal to n c, bisect it in e ; draw e c and d g cutting each other 
at m, join m c, bisect it by a perpendicular meeting c g, produced at h ; draw h d, cutting b a 
at k, and make n i equal to n k ; nl equal to ?i h; through the points i, I, k, h, draw the 
lines h i, k I, and i I, h 7c, then describe the four sectors by help of the centres, i, I, h, h, and 
it will be the representation required. 

Fig. 5. To describe an Ellipsis by Ordinates. 

Make a semicircle on the length a b, divide it into any number of equal parts, as 16, on 
the end at a make a 8 perpendicular, equal to half the width, and draw the ordinates through 
all the points in the semicircle, draw the line 8, 1 to the centre, then a 1 8 will be a scale to 
set your oval off; take 1 1 from the scale, and set it from 1 to 1 in your oval both ways at 
each end ; then take 1 2 in your scale, and set it to 1 2 in the oval, and find all the other 
points in the same manner; a curve being traced through these points will be the true 
ellipsis. 



16 PRACTICAL GEOMETRY. 



PLATE 6. 

Fig. 1. To make an Ellipsis with a String. 

Take the half a g of the longest diameter a b, and with that distance fix the foot of the compass in c, 
cross a b at e f in which stick two nails or brads, then lay a string round ef and c, fix a pencil at c, and 
move your hand round, keeping the string tight, will describe the ellipsis. 

Fig. 2. To describe an Ellipsis by a Trammel. 

1 2 3 is a trammel rod: at 1 is a nut with a hole to hold a pencil ; at 2 and 3 are two other sliding 
nuts ; make the distance of 2 from 1, half the shortest diameter of your ellipsis, and from the nut 1 to 3 
equal to half the longest, the points 2 and 3 being put into the grooves of the same size, move your pencil 
round at 1, and it will describe the true curve of an ellipsis. 

Fig. 3. Jin Ellipsis being given, to find the Centre and two Axes. 

Draw any two parallel lines a b and c d at pleasure, divide each of them in two equal parts at the 
points e andf and through ef, draw the line k I, divide k I into two equal parts at the point g - , place the 
foot of the compass in g, with the other foot make two crosses h and i, on the circumference ; draw a line 
h i, through g, draw m n parallel to h i, also through g draw o p, at right angles to m n; then o p is the 
transverse axis, and mn the conjugate, and g the centre of an ellipsis. 

Fig. 4. How to proportionate one Ellipsis within another; that is, to give it the same Length in Proportion to 

its Width, as the Length of the other has to its Width. 

Let the given ellipsis bead be, make the parallelogram e hfg to touch the sides and ends of the 
ellipsis, draw the diagonals ef and g h, of the rectangle, let r a be the width of the lesser ellipsis given, 
through the point q, or r, draw I o, or m n, parallel to the transverse axis, at the points m and n, where it 
cuts the diagonal, draw m I and n o parallel to the conjugate axis, will also show its length. 

Fig. 5. How to describe an Ellipsis about a Parallelogram, to have the same Length in Proportion to its Width, 

as the Length of the Parallelogram has to its Width. 

Let the given parallelogram be ab c d; let the diagonals a c, and b d, be drawn from the centre i; 
draw the quarter of a circle, 2 1 k, to half the width of a rectangle ; divide the quadrant into two equal 
parts at 1 ; through the point 1, draw the line I 3 parallel to the transverse axis to cut the diagonal b d in 
the point 3; then draw the lines 3 2 and 3 4; again, draw fd parallel to 2 3, then i/will be half the width, 
and d e parallel to 3 4 ; and i e will be half the length of the ellipsis: make i h equal to i e, and ig equal to 
if which will give the four points through which the ellipsis must pass ; describe the curve, and the thing 
will be done. 

Fig. 6. To divide a Line in the same Proportion as another is divided. 

d a is a line given already divided, and d e is a line to be divided in the same proportion, making any 
angle at d join a e, draw bf and c g parallel to a e; then d e is divided atf and g in the ratio of a d at b 
and c. 

Fig. 7. To do the same by an equilateral Triangle. 

a b is the given line divided : from c take two equal distances c d, e d, and by drawing lines from the 
several points in a b to c cutting d d, d d will be divided as a b. 

Fig. 8. To make an Octagon the nearest Way from a Square. 

Draw the diagonals of the square to cross at e, fix the foot of your compass in c, and take the distance 
ce and make an archfeg; then set your gauge to d f ox. b g, which will gauge off each angle. 



Plate 6. 



Fw.l. 



Fiq. 2. 





Fia. 4. 



Fit 





Fig. 5. 
f 





a 




2 


~ r7 ~~\^ 


<J 




\j 


^v 3/ 




( 




JSS 


4 \ 




















C 



I'm. 6. 



=Q 





Fia. 8. 




I, „- 




Plate. 7. 




Conic- Section 



The Ellipsis 




0^ 





1 2 3 4 5 



The Pafahola- 
i 





54321012345 



PRACTICAL GEOMETRY. 17 



PLATE 7. 

CONIC SECTIONS BY INTERSECTING LINES. 

DEFINITIONS. 

1. A cone is a solid having a circular base, from which the sides continually diminish in straight lines to a 

point in which they all terminate, and this point is called a vertex. 

2. Opposite cone is another cone joining the vertex of the other cone, with its sides everywhere in the same straight 

line passing through the vertex as a common point. 

3. A right line joining the vertex and the centre of the base is called the axis. 

4. If a cone be cut by a plane passing entirely through its curved surface, but not parallel to the base nor to 

the axis, nor to a plane touching the side of the cone, the section is an ellipsis, excepting in one position 
which is a circle. 

5. If a cone be cut by a plane parallel to its sides, the section is a parabola. 

6. If the cone be cut by a plane passing through the opposite cone, the figure will be a hyperbola. 



To describe the Ellipsis from the Cone. 

Figure A. Let B be half the circle of the base of the cone, n the vertex at the top ; then n a and nd 
are two sides ; let the cone be cut by a plane passing through g h; bisect g h at the point k, and through k 
draw r q, parallel to the base a d ; also, bisect r q in m, describe the semicircle r p q, draw kp at right 
angles to q r ; and g h is the length of the ellipsis, and h k half its width; from which the figure may be 
described at C, as explained in the next plate. 

To describe tlie Parabola from the Cone. 

Figure A. Let i e be the axis of the parabola, parallel to the other side n d of the cone, and through 
e draw e c at right angles to the base ; then will e c be half the width of the parabola, and e i its height ; 
then the figure will be described, as at D, by intersecting lines upon each ordinate, up to the crown, from 
the equal divisions on each side. 

To describe the Hyperbola from the Cone. 

Figure A. Let the axis of the hyperbola be if, cut by a plane passing throughy*and i, till it cut the 
opposite cone at I; draw/6 at right angles to a d, then is fi the height of a hyperbola, andy6 half the 
width of the base, and i I its transverse axis ; then make/* i at E equal toy i in figure A, make i I in E equal 
to i I in figure A, b b in E equal to twice fb in figure A ; let the base b b in E be divided into ten equal 
parts, as at 1 2 3 4 5, that is, into five equal parts on each side from the centre, and draw lines to the point 
I through these points; likewise divide the height into five each way, and draw lines to the vertex at i; 
this will show the points through which the curve must pass. 

3 



18 PRACTICAL GEOMETRY. 

PLATE 8. 

How to draw any Semi-ellipsis upon the transverse or conjugate Axis, or even a Semicircle 

itself, by a new Method of intersecting Lines. 

Figures A and B. Let the given axis be a b, and let it be divided into any number of 
parts, as 10 ; also let the height be divided into half the number of parts ; make e d equal 
e c, that is, to the height of the arch ; then, from the point d, draw lines through the equal 
divisions of the axis a b; likewise, through the points 1, 2, 3, 4, 5, in the height af; draw 
lines tending to the vertex at c, which will intersect at the points h, i, h, I; and lines being 
drawn through the divisions of b g to c, at the crown in the same manner, will give the 
points n, o, p, q; a curve being traced through these points, will form the true curve of an 
ellipsis. 

The semicircle, figure C, is drawn in the same manner, by making af equal to one half 
of a b. 

How to draw the true Segment of a Circle, by the Method of intersecting Lines. 

Figure D. Let a b be the length of the segment, and o c its height, and draw the 
chord b c for one half of the segment, and draw b m at right angles to b c ; and from the 
point o divide a b each way, into five equal parts ; also from c, divide c m, and c n, each 
into five equal parts; and draw 1 1, 2 2, 3 3, 4 4, 5 5, on each side, through the divisions 
1, 2, 3, 4, 5, on a s, and 1, 2, 3, 4, 5, on b r; draw lines to c, which will intersect the other 
lines at the points d, e,f g, and h, i, k, I: the curve being traced, the thing is done. 

How to draw a flat Segment of a Circle nearly true. 

Divide the length of the segment into equal parts each way, from the centre d, as 
before, and draw the lines 1 1, 2 2, 3 3, 4 4, 5 5, all at right angles, to the length a b; lines 
being drawn to the crown at c, from the divisions at each end, will show the points which 
the segment must pass through ; the curve being traced, the thing is done. 

Remark. Although this last method is not the true segment of a circle, but a para- 
bolic curve, yet it will be found useful in practice, in tracing any segment whose height 
is not more than one-tenth part of its length; if the centre of the segment is found, and 
drawn with a compass, the difference will hardly be visible, and the flatter the segment, 
this difference will become the more imperceptible ; but if the height exceeds one-tenth 
of its length, the difference will be visible ; for then the arch will be quicker at the ver- 
tex, and get flatter and flatter towards each extreme. 

In the same manner may all kinds of rampant ellipses be described, or any segment 
of them, as at F and G, also a rampant parabola in the same manner, as H. 




ft s 



Fig. D. 




Fig. E. 




Fiq.M. 




Plate 9. 



N?l Fig. 2. 




PRACTICAL GEOMETRY. 19 



PLATE 9. 

THE SECTIONS OF A CYLINDER. 

DEFINITIONS. 

A cylinder is a figure generated by the revolution of a right-angled parallelogram about one of its sides ; con- 
sequently the ends of the cylinder are equal circles, and the line passing through the centre of the cylinder 
is called the axis. 

Tlie section of a cylinder, cut by any plane, is an ellipsis, or circle, or rectangle, proved by the writers on Conic 
Sections. 

To find the Section of a Semicy Under, by Ordinates, when it is cut at right Angles to the Plane passing through 

its Axis, in the Direction a b. Fig. 1. 

Let the circle of the base be divided into equal parts at A, and drawn parallel up the cylinder to the 
line a b, at the points 0, 1, 2, 3, 4, 5, &c, and from these points draw lines at right angles to a b; then B 
being pricked from A, as the figures direct, B will be the section of the cylinder. 

DEMONSTRATION. 

Conceive the semicircle A at the base to be turned at right angles to the plane, also the semi-ellipsis 
B at right angles to the same plane ; then will the ordinates of B be parallel and perpendicular over the 
ordinates of A, and every corresponding point in the circumference of B will fall perpendicular to the same 
corresponding points in A: therefore B is the true section of the cylinder, cut in this position. 

To cut a Cylinder in the Direction a b, upon a Plane, passing through its Axis, to make an acute Angle with 

the Plane. Fig. 2. 

Let C, at No. 1, be the given angle, which the section at B is to make with the plane of the cylinder; 
take a b in figure 2, that is, the radius of the base, and set it from b c, at No. 1, perpendicular to i b; draw 
c c parallel to i b, also from c draw c e perpendicular to i b ; then take the distance c i, set it from i tof in 
figure 2, at B; likewise take i e from No. 1, and set it from i to e in figure 2, at B; draw e d parallel to m 
n, to cut the rake in d, and join df; then is df the bevel of the first ordinate of the section B. And draw 
the lines e c and d a parallel to the axis ; join a c at A; then will a c be the bevel of the first ordinate of 
the base. Draw all the other ordinates of A parallel to a c, and at the points 1, 2, 3, 4, &c. in m n, draw 
lines parallel to the axis of the cylinder, to cut the raking line V W at 1, 2, 3, 4, 5, &c. From these points 
let lines be drawn parallel to df; then the ordinates of B, being pricked from the same corresponding ordi- 
nates of the base at A, will give the section of the cylinder. 

Note. The point/ will fall beyond the sweep at the section B. 

DEMONSTRATION. 

Let the plane B be conceived to be turned round the line V W to make an angle at the point i, with 
the plane nm V IT equal to the angle eic, No. 1 ; and conceive a straight line drawn from e perpendicular 



20 PRACTICAL GEOMETRY. 

to the plane n m V W, the line thus supposed to be drawn "will be parallel to the plane A of the base, and 
the triangle formed by if, i c, and the perpendicular from e, will be equal and similar to the triangle c i e, 
No. 1 : then because d e and the perpendicular are both parallel to the base, the line that joins the points e 
andy will also be parallel to the base; and because e d is equal to a 6, the triangle e df will be equal and 
similar to the triangle 6 a c in the plane of the base A: and because 6 c and the perpendicular drawn from 
e are both in a plane parallel to the axis, the plane passing through df and a c will also be parallel to the 
axis : but df is also in the plane of the section, for the point d is the intersection V W, and the point f 
will be in the perpendicular drawn from e ; therefore, if a series of planes be conceived to be drawn through 
the ordinates of the base parallel to the plane passing through a c and df, the intersection of these planes 
on the plane of section B will be parallel to the ordinates of B, and every two corresponding lines will be 
in a plane parallel to the axis, and therefore as the lines formed by the intersections of the series of planes 
in the section B, are equal to those in the base A, the extremities are in the section of the cylinder. 

To cut a Segment of a Cylinder, in the Direction a b, to make an obtuse Angle with the Plane of the Segment. 

Fig. 3. 

Let No 1 be the angle given, which the section B is to make with the plane of the segment ; from f 
in No. 1, drawyg- at right angles to fc, and g e also perpendicular, to make the right-angled triangle e gf. 
And in figure 3, at B, draw gf, at right angles to a b, and make g e equal to g e at No. 1. Also, make 
gfatB equal to gf&t No. 1. Draw e d at B, parallel to M JY, and at the point d, where it intersects the 
line a b, join df; then df is one of the ordinates. From e and d, draw the two parallel lines e c and d 3, join 
c 3; then c 3 will also be an ordinate of the base. Draw parallel lines at discretion to c 3, for the. other 
ordinates of the base ; and from their intersection upon m n draw lines parallel upon the cylinder, to cut a b 
in 1,2, 3, 4, &c, and from these points draw parallel lines to df, which are the ordinates of B; these, being 
pricked from the base as the figures direct, will give the points through which the curve must pass, which 
being traced, will be the true section of the segment of the cylinder. 

DEMONSTRATION 
Is the same as the preceding Demonstration. 

That the reader may perceive this more clearly, the best way is to draw those lines on pasteboard. 
The section and the end being made to turn round, in their proper position, then the demonstration will be 
clearly seen. 

Figure 4 is to be laid down and demonstrated in the same manner as Figure 2. 

Remark. Upon these figures depend the whole principles of hand-rails for stairs. The reader ought 
to understand how to form the section of a cylinder, in any case whatever ; for the face of raking mould 
of a hand-rail is nothing but the double section of a cylinder, as in figure 4, at B, where the double circle 
upon the base A represents the plan of a rail, and the bevel at No. 1, figure 4, represents the spring of the 
plank, and a b the pitch of the rail : therefore, it is very necessary that the reader should have a knowledge 
of these figures and their demonstrations ; and not be satisfied with only doing it, but read these demon- 
strations, and consider them with attention, then he will be able to see the reason why every line is drawn 
in the manner it is. 



Furl 



Plate 10. 



Fig. 2. 




Fir/. 4. 



Fig. 5. 




5 4 3 2 1 



PRACTICAL GEOMETRY. 21 

PLATE 10. 

THE SECTIONS OF A GLOBE, OR ANY OTHER FIGURE STANDING UPON A CIRCU- 
LAR BASE: ALSO, THE SECTION OF ANY FIGURE STANDING ON AN IRREGULAR 
BASE. 

DEFINITION. 

A globe is a figure generated by the revolution of a semicircle round its diameter, which becomes the axis of 

the globe. 

AXIOMS: OR, SELF-EVIDENT TRUTHS. 

1st. From this definition it appears, that every two sections passing through the centre, are equal to each 
other. 

2d. Every section of a globe, cut by a plane, is a circle; for the generating circle may be made to revolve 
round any line, as an axis ; and therefore every point in it will generate a circle, whose diameter must be 
twice the radius of that circle distant from the axis of the globe. 

3d. If a semi-globe is cut by a plane at right angles to the plane of its base, the section will be a semicircle. 



To find the Section of a Semi-globe cut by a Plane at right Angles to the Plane of its Base. Fig. 1. 

It appears from the last axiom, that there is no tracing required : for, let the section be cut across a b, 
figure 1 ; divide a 6 in two equal parts at the point c; and on c, as a centre with the radius c a or cb, describe 
the semicircle A, which is the true section required. 

The same Ordinates. Fig. 2. 

Draw any line d e through its centre, and let a b be the place of the section upon the base, as before ; 
place the foot of your compass in the centre of the globe aty, and, with the radius t c, draw an arch from 
c to g, in the diameter d e ; the foot of your compass remaining still iny, draw the concentric dotted circles 
from c b tofe, and at the intersecting points 1 2 3 4 5 in f c, and likewise in c b, erect perpendiculars to 
those lines ; then A being pricked from C, as the figures direct, will give the points through which this se- 
micircle must pass. 

DEMONSTRATION. 

Conceive the semicircle C to stand at right angles upon d e, also the section A to be at right angles to 
a b ; now it is evident if g i is the height of the globe over the point g in the base, c 1, which is equal to 
g 1, must also be the height of the section, because the points c and g stand at an equal distance from the 
centre ; and therefore the point 1 over c, is in the surface of the globe. In the same manner it may be proved, 
that any other points carried round by the dotted lines are in the same surface ; but the section that stands 
upon a b, in A, is a semicircle ; and consequently the method of tracing is also a semicircle. 



22 PRACTICAL GEOMETRY. 

Observation. Hence appears the erroneous principle of tracing used by a late writer upon this subject, 
as you may see at figure 2, where A is the section of a globe, and the bracket at D is the section across the 
diameter. A is truly traced from D, because the ordinates are carried round in circles ; but by his method 
of tracing, as you see at C, upon the other side, the point of the bracket C falls within the sweep of the 
circle, by reason of the ordinate of C being carried straight through between the two bases, which I have 
proved to be false. And this he has applied in bracketing up the angles in the square well hole of a stair 
case, to the circular curb of a sky-light, which if truly done, is nothing else than upon the same principle as 
the sections of a globe. 

Figure 3 is done upon the same principle as figure 1. A is the section traced from C, and wants no 
other demonstration than what has been given in figure 1. 

Figure 4 is an ogee section, standing upon a circular base across the diameter ; and A is the section 
traced from it, upon the same principles as figure 1. 

From these examples it is clear that this method of tracing does not depend on the form of the top, 
but entirely upon the base. These figures are supposed to be generated round an axis ; and, as every circle 
is carried round at an equal distance from the axis, the perpendicular height of the figure, upon any circle, 
must be the same height in every point throughout the circle : which proves itself to be the only method 
for anything of this kind. 

A Semi-globe being cut by a cylindrical Surf ace perpendicular to the Plane of its Base, to find the Form of a 

Veneer that will bend round it. Fig. 5. 

Let d e be drawn through the centre jf; and place the foot of your compass in/, the centre; and draw 
the points 6, 1, 2, 3, 4, which are equally divided from the centre at b, in the circular surface, draw the 
concentric dotted lines round to the diameter d e, at 0, 1, 2, 3, 4, and at these points raise the perpendiculars 
0, 1 1, 2 2, 3 3, 4 4. Take the stretch-out round 612 3 4 5, which is one-half; and lay it upon the base 
of No. 1, each way, from 12 3 4, &c, and No. 1 being pricked from A, figure 5, as the figures direct, 
will give the points through which the curve must pass for the veneer. 

DEMONSTRATION. 

For, since the section standing upon d e is a semicircle, which is equal to the semicircle upon the 
base; and as the points 1 2 3 4 in the circular surface, stand at the same distance from the centre f as 
0, 1, 2, 3, 4, in d e; now if the point o at No. 1, is made to coincide with the point b in figure 5, then the 
height o o, standing over the point b, will be equal to the height o o at A ; but these points are at an 
equal distance from the centre, therefore the top of each ordinate will be in the surface of the globe. In the 
same manner every other point may be proved, when bent round and elevated, to be of the same height, 
and at an equal distance from the centre with those of A; and therefore No. 1 is the true form of the 
veneer. 

To find the Ribs of a Gothic JYiche, being the Plan, and No. 1, the Front Elevation. Fig. 6. 

Take the length of each base upon the plan, and make them the bases of No. 2, No. 3, No. 4, and 
No. 5 ; divide each base into five equal parts ; also divide the half of No. 1, into six parts, and draw the 
ordinates from the equal divisions, perpendicular to each base ; then prick each from No. 1, as the figures 
direct, will give the form of each rib. This wants no demonstration. 



Late 1J. 

Liirino for a Pcirallel soffi itting oblirfzw 
l?i a straight inn,' 

S 7. . c7i out line 



{e /fiJ ati p ■/>,, 




Lining fen a Parallel soffit cutting 

R ' , ; /'■ / / / / ,- , ' t / < v / a/ lav lit/// 

fni.Y. 






Lmiih/ for it Parallel soffit ciUtinq 

Oolique into a. Circular* wall 

Fuji 




Stretch oat Jive 

jo .9 . S. 7 



OF CARPENTRY. 



LININGS FOR SOFFITS. 

DEFINITION. 

The lining of a Soffit, in the Theory of Carpentry, signifies the covering of any concave Surface of a Solid 

spread out on a Plane if possible. 

Soffit in Architecture is the under side of the head of a door, window, or the intrados of an arch, and may be 

either plane or curved. 



PLATE 11. 

How to stretch out a Soffit, when a Window or Door, having a semicircular Head, cuts into a straight Wall, 

in an oblique Direction. 

LET C be the plan or opening of the window, in fig. A, and let the base of the semicircle B be drawn 
at right angles to the jambs, or sides of the plan C ; divide the semicircle into any number of equal parts, 
as ten, and draw the ordinates across the plan, extend the parts round B upon the stretch-out line, the or- 
dinates being drawn from the divisions across, and traced off from the plan C, as the figures and letters 
direct, the lining of the soffit will then be completed. 

If you would make a cylinder to be only the thickness of the wall, D shows the end of it, which is to 
be traced from the semicircle B. 

How to draw the lining of a Soffit when the Top is a Semicircle, cutting right into a circular Wall. 

Fig. E. This and the other below are performed the same as that above, with this difference, that you 
are to prick from the circular plan, instead of the straight plan. 

Fig. 1 shows the method when a circular headed window cuts oblique into a circular wall. 

Note. In all kinds of cylindro-cylindric soffits, when the two jambs are parallel, the straight line, 
which the soffit is pricked from, must be drawn at right angles to the jambs, as is shown in this plate ; for 
want of this consideration, they are shown in books upon wrong principles. 

But in the following soffits, where the jambs are not parallel, they must be continued till they meet in 
a point, and the line which the soffit is to be pricked from, must be made to form an isosceles triangle wilh 
the jambs. 



24 LININGS FOR SOFFITS. 



PLATE 12. 

To draw the lining of a Soffit in a straight Wall, splaying equally all round with a circular 

Head. 

In fig. A, continue the sides of the plan A, that is a c and b d, to meet at e ; then about 
the centre e, and from the points a and c, describe the soffit C, and stretch the semicircle B 
along the outline of the soffit C, it will be completed. 

To draw the lining of a Soffit in a circular Wall, splaying equally all round with a circular 

Head. 

Fig. B. The stretch-out of this soffit is managed the same as in the last ; draw the 
ordinates of the semicircle B, from thence continue them to f the concourse of the splay, 
and at the points a, b, c, d, e, where they intersect the plan, draw the parallel lines a e, bf 
c g, &c. parallel to the base of B, and from the points e,fg, h and i, circle lines to a, b, c, d 
and e, round the centre^ which will give the half of one edge of the soffit, the other half 
being pricked from it ; the other edge is found in the same manner. 

Note. This cannot be pricked from the plan as the others are, as the lines round the 
splay are not level with the plan, and will therefore be longer than those on the plan. 

DEMONSTRATION of Fig. A. 

Conceive the semicircle B to be turned at right angles to the plan A, then every point 
in the circumference of the semicircle B will be at an equal distance from the point e, but 
the soffit C is described with the same radius ; therefore the edge of the soffit C, that is, the 
arch line af, will exactly coincide with the arch of the semicircle B, which was to be proved. 

DEMONSTRATION of Fig. B. 

It is easy to conceive from the last demonstration, that if the semicircle B is turned up, 
and the soffit at C bent round it, the points 1, 2, 3, 4, 5, at C, will coincide with equal divi- 
sions imthe semicircle B, and the points a, b, c, d, &c. at C, will fall perpendicularly over 
the points abed, &c. in r the plan A : for the arches a e, bf, c g, d k, and e i at C, will fall 
over the parallel straight lines e a,f b, g c, h d, i e, in the plan A, which was to be demon- 
strated. 

The learner is advised to cut these and the following soffits out of pasteboard, and their 
demonstrations will be more clearly seen. 



Lining for 
,i soffit cutting right 
i7i/<> a straight wall ^ v \> v 

Fining equally till round. ^" 

Fig.: V 




Lining tor 
a soffit in 1 1 C 1 1 
Wait fizzing >\/u> 



Plate 23. 




LININGS FOR SOFFITS. 25 



PLATE 13. 

To find the lining of an Aperture whose Plan A BCD is a Trapezoid, with two parallel 
Sides A B and D C, which represent the out and insides of the Wall, and two equally in- 
clined Sides A D and B C, which represent the Jambs, and whose elevation A I B, on the 
inside of the Wall is a semi-ellipsis, and that on the outside DG Ca Semicircle, so that 
a straight Edge may everywhere coincide with the lined Surface, and be parallel to the 
Horizon. 

Produce A D and B C to E; bisect D C at F, and draw E F; produce it to I and it 
will cut A B at H; then F G and ill are equal to each other. Divide the quadrant D G 
into any number of equal parts, (as five,) and draw lines through the points of divisions cut- 
ting the base D C ; from the points of division and in the same straight line with the point 
E, draw lines to cut A H, and the lines so intercepted will represent the level straight lines 
on the soffit. Make E J perpendicular to E D equal to F G, and mark the other divisions 
on E J from E, at a, b, c, d, respectively equal to those in F D : then take the distances of 
the several points in D C from E, beginning next E D, and proceeding to the last E F, and 
describe the arcs from the centres, a, b, c, d, respectively ; with the fifth part of the arc D G 
fix the foot of the compass in D, and cross the first arc at c ; place one foot of the compass 
in e, cross the next arc aXf proceed in this manner to i, then D, e,f, g, h, i, will be the coin- 
cident line of the lining or interior covering for the arc D G. Join i J and produce it to 
K; make the angle KJL equal to the angle KJE, and make / L equal to J E: mark 
the divisions on J L, so that the distances from J" may be equal to the distances of the 
several divisions on J E, then the other half of plano-cunioidal line may be found by in- 
version. Produce the lines a e, bf, c g, d h,j i, &c, to o, p, q, r, s, &c, make e o, fp, g q, 
&c, in inverted order equal to the seats of the lines on the soffits and the points o, p, 
q, r, &c, then curves being drawn through the points o, p, q, r, s, &c, and through e, f, 
g, h, &c, will form the wall lines of the covering, so that AD MN will be the whole 
covering or interior development. 



26 LININGS FOR SOFFITS. 



PLATE 14. 

To find the lining of an Aperture covered with the same Surface, and terminated by a 

circular Wall. 

Find the line D Q M, as in the last plate, as in a straight wall, then transfer the dis- 
tances from the straight line on the plan to the arcs, representing the faces of the wall to the 
covering, and the edges will be obtained as in the last. 



Plate //. 




Plate 15. 



Fig. A. 





l TO 3 





LININGS FOR SOFFITS AND TRACERY. 27 



PLATE 15. 

To draw the lining of a cylindrical Soffit, cutting right in a Wall which does not stand 
perpendicular to the ground to a level Base. Fig. A. 

Let a e at D be the level of the ground, a I the inclination of the wall, equal to the 
radius of the cylinder ; let fall the perpendicular from I to c, in the bottom line a e make the 
semicircle in fig. A ; to the width of the cylinder, or the double of a I at D, take the dis- 
tance a c at D, and make b a equal to it in fig. A, and describe a semi-ellipsis to the length 
of the semicircle d d and to the width a b ; lay the equal divisions round the semicircle in 
fig. A in C, along the line d d, on each side of the middle point 4, then take the parts e d, 
d c, c b, b a, from the plan B, and lay them at D respectively from e towards d, c, b, and from 
I draw I e to make a right angle with I a, and at the points a, b, c, d erect perpendiculars to 
a e to cut I e at/*, g, h and i : take the distances e i, i h, h g, and gfi, in D, and lay them on 
the soffit at C respectively, from 1 d, 2 c, 3 b, 4 a, each way, then will the straight line d d 
in the soffit, when bent round, be perpendicular over the elliptic line in the plan B, and the 
curve^line d d c b a, &c, d will fall over the points d d c b a in the plan : in the same manner 
the edge of the soffit may be brought to answer any curve line proposed. 

To draw the Arches of Groins by a new Method, whether right or ra?npant, so that their 
Arches shall intersect or mitre truly together, from a given Arch of any Form. 

Let fig. E be the given arch of a Gothic form, draw the chord a c for one-half the arch, 
divide it into any number of parts, as four, and through the divisions draw lines from the 
centre e to terminate in the circumference at h, g, I, draw lines from c through hglto cut 
the perpendicular a d at b, c, d ; and if No. 2 is required to be wider, but the same height as 
fig. E, draw the two chords a c and c b for each side of the arch, divide each into four equal 
parts, as before, and set the parts a b, b c, c d perpendicular on each end of a b at No. 2, 
and from the divisions draw lines to the vertex at c, then trace the curve through the points 
h, g, I, &c, so the arch at No. 2 will truly mitre into fig. E; in the same manner the ram- 
pant curve at No. 3 will be brought to correspond with fig. E, and No. 2.* 

* It is hardly possible to find a more ready method in practice, because a chalk line will soon strike all the 
radical lines, having only to move it but once from the point e up to c at the crown ; Jig. F. shows the common 
method by dividing the basis of each into a like number of parts, and transferring the height, as the figures explain, 
at No. 1 and No. 2; nothing is more tedious in practice than raising a number of large perpendiculars, and going 
continually from one curve to get the height of another. 



28 CHURCH WORK AND RANGING OF RIBS. 



PLATE 16. 

./2s it happens sometimes in church work, that windows go higher than the ceiling line, which therefore requires 
to be hollowed out, so that the light may be thrown down into the body of the church: I shall in this place 
show the method of making a curb for that purpose. 

To Find the Form of the Curb. 

Let k b I be the head of the window, figure A, and let it come as high as a b, above the ceiling ;* and 
let a b at No. 1, be the same height, and b c the direction of the light, and a c will be the length of the curb. 
Make a c at No. 2, equal to a c at No. 1, and divide it into six equal parts; also divide a b, in figure A, 
into six equal parts, and let the ordinates be drawn as is explained in the figure ; a curve being traced round 
the points of intersection, will give the form of the curb. 

Figures B and C show the method of drawing and backing any elliptic rib with a compass, which is 
exceedingly handy in drawing, and will be near enough for the representation of an elliptic rib on paper, 
as no other method will be so clean when done ; but for practice, a trammel, or intersecting lines is more 
ready. 

To draw and range the Ribs by this Method. 

In B, let c h be the height, and c b the width ; divide the difference into three equal parts, and set four 
such parts on each side of c, to d and d, and make an intersection with the distance d d at e, and draw a 
line through e and d to i, then d and e are the centres for the interior side : suppose the rib is to be backed 
as much as a b upon the bottom, set a b from d tof, and from e to g, parallel to the base; and draw a line 
through g,f, to k: then g andf are the centres for describing the ranging lines. 

The rib E is traced from D, and a b being set all round on the parallel lines, shows how the ranging 
is found for a drawing on paper : the rib C is described in the same manner. 

The word backing is properly applied to the upper side of anything in Carpentry or Joinery, as the 
back of a rafter, the back of a rail ; but range applies either to the operation of levelling the upper or 
lower edge, and explains its own meaning, viz. forming the edge so as to range with the other edges, whe- 
ther forming a ceiling or the exterior of a roof. 

* The ceiling is here supposed to be level, which is seldom the case in a church ; hut the method will be no- 
thing different if the ceiling line a e were to incline to the horizon in any angle whatever, only observe to make 
cab equal to that angle. 




Ceiling line upon the transverse 



Ceil in a upon 




9 <• 



Plate 17. 




CENTRINGS TO GROINS. 29 

PLATE 17. 

DEFINITION. 

Groins are formed by the intersection of arches or vaults, and the surfaces of their meeting may be considered 

as the sections of cylinders, cylindroids, fyc. 

BRICK GROWS. Description. 

P, P, P, &c, is the plan of the piers which the vault is to stand upon, a b,fig. D, is the end opening, 
which is a given semicircle; and b c is the opening of the side arch, which is to come to the same height 
as the end arch a b: fix your centres over the body range, fig. A, as shown in the section at C, then board 
them over. In fig. A, is the manner of fixing the jack ribs upon the boards, which likewise shows at C. 

To find the Mould for the Jack Ribs. 

Take the opening of your arches in fig. A, that is, a b and a d, and lay them down in fig. D, at a b 
and b c, to make a right angle. Divide one-half of the given semicircle into five parts, and square them 
across 1, 1, 1, &c, to cut b d and d c, the diagonals in 2, 2, 2, &c, and through the points 2, 2, 2, &c, 
draw lines parallel to 1,1, 1, &c, the base of E, both ways towards F and G ; stick in nails at 1, 2, 3, 4, 5, 
in the arc of E, and bend a thin slip of wood round them, which mark with a pencil at every nail ; this slip 
of wood being stretched out from d, 1,2, 3, 4, 5, and squared over to G, will intersect the other lines in 
small rectangles : a curve being traced through the diagonals of each rectangle, will give a mould to set the 
jack ribs. 

How to fix the Jack Ribs. 

Bend your mould G from a, to the crown at e, in fig. A, that will give the edges of your boards ; then 
fix a temporary piece of wood, level upon the crown, in the direction offf and let it come the thick- 
ness of your boards lower than the crown, and it will give the height of your jack ribs, which is a very 
sure method of placing them. 

To find a Mould to cut the ends of the Boards. 

The rib F is traced to the height of E, or got by a trammel, which will be fully exemplified in the 
following plates. Take the parts round F, and lay them out to 1, 2, 3, 4, 5; then if will be got in the 
same manner as G, which will be a mould to cut the ends of your board that goes upon the jack ribs 
against the body range. 

Fig. 1. Is an easy method of getting the moulds when both arches are the same opening. 

Take half the opening of the arches, whatever they are, and draw a quarter circle, and divide it into 
six ; bend a slip round it to take its parts, then stretch it out upon the base from to 6, and square over 
your points 1, 2, 3, &c. Through the points in the arch draw the lines on both sides parallel to 0, 6, the 
curve being traced as before, gives both moulds of an equal and similar form. 

Note. The curve Pmay be drawn in practice with a trammel, independent of the other, and the two 
moulds Fand G may be drawn separate, without any connection of lines, as shall be shown hereafter. 



30 CRADLING TO GROINS. 



PLATE 18. 

DEFINITIONS. 
Groins are said to be ascending or descending when they are not built upon level ground. 

CENTRING FOR ASCENDING OR DESCENDING GROINS. 

The Plan and Inclination of any Groin being given, and one of the Body Ribs, as B, also the Place of the 
Jingles upon the Plan, to find the Form of the Side Ribs, so that the Intersection of both Arches will be 
perpendicularly over the Plan. 

Divide half the circumference of the given rib B, into any number of equal parts, and draw them to 
intersect the angles ; and from thence let them be returned up to the rib C, upon the side ; then C being 
pricked from the given rib at B, as the letters direct, will give the form of the side centre. The same is 
shown at F, by the method of intersecting lines. 

To find the two moulds D and Efor placing the Jack Ribs, to bend over the Jingles in the Body Range, when 
boarded in, so that they may be perpendicular over the Jingles upon the Plan. 

At C, draw lines from the points a, b, c, d, e,f, g, &c, where the ordinates of C intersect the top of 
the arch, perpendicular to the rake, and draw the semi-ellipsis, Ji, to the width of the body range ; and to 
a h, the height of the side centres, perpendicular to the rake ; and continue the ordinates of B, up to Ji, 
to intersect at 1, 2, 3, 4, 5, 6. Bend a slip round these points, and mark them opposite to every point, and 
stretch it out along k, 1, 2, 3, 4, 5, 6, between D and E, and draw lines through these points, at right 
angles to k 6, to intersect with the perpendiculars. Begin at 6, and trace a curve both ways, will give the 
edges of the two moulds for placing the jack ribs. 

To cut the Jack Ribs to the Rake of the Groins. 

Set the number of the jack ribs upon the arch B, at their proper distances, and take their several 
heights, that is, h i, k I, m n, and set them upon the arch G, from a to 6, and from a to c, and from a to d, 
draw lines through these points parallel to the rake, which will show how the jack ribs are to be cut, so 
that they shall range properly with the other raking centres. 

Note. All the body ribs must be ranged according to the rake of the groin ; to do this exactly, the 
under edges of all the ribs must be bevelled according to the rake ; then make a mould as B, or one of the 
body ribs themselves will answer instead of a mould, which being applied to each side of any other rib, 
keeping the bottom fair with the under edge upon each side, and drawing the curves by the other, it will 
give the ranging line. 



Plate 18. 



A 5 




Plate 19. 




CRADLING TO GROINS. 31 



PLATE 19. 

Given the two side Arches of any Groin, and the Inclination, to find the Intersection of the 

Angles upon the Plan. 

Divide half of the body rib B into equal parts, and draw parallel lines to b, c, d, e and/; 
and from the point a, as a centre, draw the concentric dotted circles round to g, h, i, k, I; 
then draw parallel lines to the rake, to cut the centre C at 1, 2, 3, 4, 5, and a, b, c, d, e, on 
the other side ; and from these points let lines be drawn perpendicular through the plan. 
And on the centre g of the rib C, square a line up to 5, the top of the arch C; and from 5 
draw a line perpendicular through the plan. Also through the points 1, 2, 3, 4, 5, at B, 
draw perpendicular lines to the plan the other way ; begin at h, and trace through the angles 
both ways, will give the place of the angles upon the plan. 

The moulds for bending over the angles are found in the same manner as in the last 
plate, by taking the stretch-out round A, and laying it between D and E. 

The reader may see such groins executed under the Adelphi buildings in the Strand, 
London, where the declivity is very rapid in going to the river. 

The jack ribs of the groin are cut in the same manner as directed in the last plate, and 
in the practice there will be no occasion for tracing the angles, as the two moulds D and E 
are done independent of them ; the reader will farther observe, that the arch B must not be 
used instead of the arch A, which would produce a very great error in the moulds D and E, 
as it must be evident to every one, that the section upon the square of the cylinder, or body 
range, must be less in the height than the perpendicular or plumb section B, which in this 
case is oblique : if these things are properly understood, there will occur nothing in brick 
groins but what may be easily surmounted. 

In all kinds of brick groins the centres or body ribs must be fixed first in the same 
manner as if there were no side arches cutting across them ; then the centres must be boarded 
over ; then to find the place of the angles upon the boards, that is, the proper intersection of 
the side arches upon the plan, the moulds D and E must be both bent round the boards at 
one time, by keeping the points I and e of the moulds D and E upon the tops of the piers 
at o and e ; then keep the top points together, and bend them round, keeping them still to- 
gether, then the point at 5 will fall perpendicularly over h in the plan ; round the inner edges 
of the moulds draw a curve upon the boards, which will be the proper intersection of the side 
arch. The jack ribs are cut in the same manner as directed in the last. 



32 CRADLING TO GROINS. 



PLATE 20. 

The Angles or Diagonals of any Plaster Groin, which are straight upon the Plan, and one of 
the Side Arches being given, to find the other Side Arch and Angle Rib. 

Fig. I, Case I. If the given rib is a semicircle or semi-ellipsis, they maybe described 
as in fig. 2, plate 6, with a trammel, which is by far the readiest method ; but if a proper 
trammel is not to be got, a temporary one may easily be made, which will answer equally 
well by fixing two pieces of wood in the form of a square, that is, to make a right angle; 
each leg must be as long as the difference between the semi-transverse and semi-conjugate 
axis, and instead of the sliding nuts in the rod, two brad awls will answer the purpose, being 
put through any straight slip of wood ; and by moving this round either the exterior or the 
interior angles of the square, keeping the pins or brad awls close to each leg, it will describe 
one quarter of an ellipsis at one time. 

To find the length of the Jack Ribs. 

Lay down the plan of the ribs as at B, and draw a rib upon each opening; then draw 
perpendicular lines from the plan of each opening, at the extremities a c e, to cut its corre- 
sponding ribs at b dfi. then the distance from b to b shows the length of the first jack rib, 
from d to d the length of the second, and from.fi toy the third. 

How to bevel the Angle Ribs, so that they shall range with each opening of the Groin. 

First get the ribs out in two halves or thicknesses, as at E and F, then draw the plan 
of your angle rib, which is placed between E and F, will show the true ranging upon the 
bottom of the rib ; then shift your hip mould parallel upon the base of E and F, will show 
how much wood there is to be bevelled off; then nail the two halves together, and it will be 
completed. 

METHOD I. 

Fig. 2. Case 2. When the given rib is a segment of a circle, or any other curve 
whatever, the ribs will be described as in plate 15, fig. E, as are shown at B, E, and F. 

method II. 

When the given arch is a segment of a circle as at A, take its height b c, and place it 
from b to c at C and D ; then take the whole diameter of the arch A, that is, twice the radius 
a c, and place it from the crown of the other arches perpendicular to their bases from c to b 
at C, and from c to d at D ; then the arch may be drawn as in plate 8, by intersecting lines : 
the ranging of the ribs is done in the same manner as in the last groin. 

Either of these two methods is much readier in practice than tracing the ribs through 
ordinates. 



Fhite 20. 



Fiq.l 




Fig. 2. 




Plate 21. 




CRADLING TO GROINS. 33 



PLATE 21/ 

Given one of the Body Ribs, and the Angles straight upon the Plan, and the ascent of a 
Groin not standing upon level Ground, to find the Form of the ascending Arches, and the 
Angle Ribs. 

Let b a c at B be the angle of the ascent, from the point b make b c perpendicular to a b, 
and describe the rampant curve B, as in plate 15, at No. 3, in fig. E: then draw the dia- 
gonal a b at E, and make b c perpendicular to it, and equal to b c at B ; then draw the hypo- 
thenuse a c, and describe the angle rib E, in the same manner as that of B. 

To find the Length of the Jack Ribs, so that they shall fit to the Rake of the Groin. 

Draw lines up from the plan to the arch, as at D, in the same manner as explained in 
the last plate ; then the arch from a to a is the first jack rib, from b to b the second, and from 
c to c the third, &c. 

How to range the Angle Ribs for such sort of Groins. 

Get the ribs out in two halves, as in the last plate, then the bottom of the ribs must be 
beveled agreeably to the ascent of the groin, and the plan of it must be drawn upon the 
level, and from thence they may be drawn perpendicular from the plan to the rake of the 
rib ; then take a mould to the form of the rib, or the rib itself, and slide this agreeably to 
the rake to the distance that is marked upon the bottom to be backed off, will show how much 
the rib is to bevel all round. 



34 CRADLING TO GROINS. 



PLATE 22. 

OF GROINS CUTTING UNDER PITCH. 



DEFINITION. 



When the side arches of a groin are lower than the body arch, then they are called under 

pitch groins. 



Given one of the Body Ribs B, and the height f g, of a Door or Window, fyc. at D, and its 
Width m 1, to find the Side and Angle Ribs D and E, so that the Intersection of the Side 
Arch D, with the Body Rib B, shall be straight upon the Plan. 

Draw c e perpendicular to c b, the base of B, and equal to the height of the window at 

D, that is, equal to fig; through e draw e a parallel to c b, cutting the arch B in a; let fall 
the perpendicular a b to c b, and continue it so as to cut the liney^ produced to k, and draw 
k m and k I, which is the place of the angles upon the plan, or the base of the angle ribs ; 
then the ribs D and E may be described from the given rib F, as directed in plate 15, fig. 

E, from a centre, or they may be described as at fig. F, of the same plate, as you see on the 
other side at A and C by ordinates ; but the first is by far the easiest method for practice, 
for if you stick a pin or brad awl in g, at D, and lay a chalk line to it, you may strike all 
the radical lines g 1, g 2, g 3, g .4, &c, in much less time than the parallel lines in A and C 
can be drawn, and with much greater accuracy ; and the divisions upon c n of the arch F, 
may be marked upon a rod, and readily transferred to the arches D and E, on mp, and fig : 
then move your brad awl out of g, and stick it in the crown at^ and strike lines from the 
divisions of m p to cross the other lines, will give the points through which the arch must 
pass ; but the reader must recollect that four or five points will not be sufficient in the prac- 
tice for tracing the curve with accuracy, and therefore a greater number must be found. At 
the other end of the groin is shown the manner in which it may be fixed, sufficiently intel- 
ligible for a workman. 



FLate .?.?. 




I'hik 23.. 




CRADLING TO GROINS. 35 

PLATE 23. 

A CYLIJYDRO-CYLIJVDRIC ARCH. 

DEFINITION. 

A cylindro-cylindric arch or Welsh groin is an under pitch groin, whose side and body arches are both given 
semicircles, or they may be similar segments of circles cutting through one another, whose intersections do 
not meet in a plane surface, that is, the place of the ribs will not be straight upon the plan, but will gene- 
rate a curve line. 



Given the Body Rib A, and the side Rib B, of a cylindro-Cylindric Arch, to find a Mould for the intersecting 

Ribs. 

Divide half the arch B, into any number of equal parts, 1, 2, 3 d, or they may be taken at discretion, 
and from these points let fall perpendiculars to a b, its base ; produce them at pleasure ; also from the same 
points 1, 2, 3, d, draw lines parallel to a b, the base of B, to intersect the perpendicular line ef\ transfer 
the divisions from e^to eg; then from the division of eg draw lines parallel to p q, to intersect the body 
rib A at the points huwy. from these points draw perpendiculars to p q, its base, and continue them to 
intersect with the perpendiculars from B, at the points k, I, m, n, between C and D ; then trace a curve 
through these points, which will be the place of the intersecting ribs upon the plan ; then draw two other 
curve lines on each side of k I m n, &c, to make the thickness of the rib upon the plan : on the inside of the 
curve draw two chords for each half to their extremities, draw two other lines parallel to them to touch the 
outside curve, then the distance between those two straight lines will show what thickness of stuff' it will 
take to make the intersecting rib ; through the points k I m n, &c, draw perpendicular lines to the chords, 
make the heights c d, 3 3, 2 2, 1 1, &c, at D, equal to the corresponding heights at B : then D is the mould 
for the intersecting rib ; C is the same as -D. 

To range the Ribs, so that they will stand perpendicular over the Plan. 

At the points x, v, t, i, in the base of C, draw the parallel dotted lines to the ordinatesof Cand D, and 
make their corresponding heights equal to those of the arch B or A ; draw the dotted curve line h u w y at 
C, and it will show how much is to be beveled off on that side of the rib; in like manner the other side D 
is beveled, as is shown by the dotted curve line. 

To find a Mould to bend under the intersecting Ribs, so that it shall give the Place of the Angle truly upon 

the Plan. 

Take the stretch round the under side of the rib D at the dots, by bending a thin slip of wood round 
it, mark it at each dot, and stretch it out along the straight line b c at E, draw the ordinates across, and 
prick them from the plan that lies between D and C, then E agreeably to the letters will be the mould re- 
quired. 

Note. The straight edge of the mould must be kept exactly to the face of the rib; when it is bent round, 
then draw a curve round the under side of the rib by the other edge of the mould, will give the true place of the 
angle. 



36 CRADLING TO GROINS. 



PLATE 24. 

There will be no occasion for explaining the lines of this groin, as they are of the same 
nature as those in the last plate ; but it will be proper to take notice, that this is a bevel groin ; 
the ribs must lie in the same direction as the plan of the groin, which will make them 
longer than their corresponding given arches at the top, but of the same height ; they are 
consequently ellipses, being the sections of cylinders ; therefore, to make a rib over I m, across 
the two piers, take the extent of the base I m, and the height of the given arch n o, and de- 
scribe an ellipsis ; and to describe the side arches between any two piers, as from a to b, take 
the extent a b, and the height of the given arch, p g, at A, and describe an ellipsis, it will give 
the proper form of the rib to stand over a b ; the intersecting ribs will require two moulds 
C and D, owing to the groins being bevel upon the plan. 

Note. The letters are marked the same upon D and C as they are upon E, to show they are traced from it. 



Plate-- 24 




Fiq . 1. 





CRADLING TO GROINS. 



PLATE 25. 

To describe the intersecting or Angle Ribs of a Groin standing upon an octagon Plan, the 
Side and Body Ribs being given both to the same Height. 

Fig. 1. £ is a given body rib, which may be either a semicircle or a semi-ellipsis, and 
A is a side rib given of the same height ; D is a rib across the angles : trace from E, the 
basis of both being divided into a like number of equal parts, divide the base of the given 
rib A, into the same number of parts ; from these points draw lines across the groin to its 
centre at m, and from the divisions of the base of the other rib D, draw lines parallel to the 
side of the groin, then trace the angle curves through the quadrilaterals, will be the place 
of the intersecting ribs ; draw the chords a b and b c, then prick the moulds B and C from 
E or D, but take care not to prick them from the crooked line at the base, but from the straight 
chords a b and b c. 

To describe and range the Angle Ribs of a Groin circular upon the Plan, the Side and Body 

Arches being given, as in the last Groin. 

The ribs are described in the same manner as in the last example for the octagon groin, 
or in the same manner as the cylindro-cylindric, Plate 23, and the ranging is found in the 
same manner as is described in that plate. 

Note. E and /"are the same moulds as are shown at B and D. 



38 CRADLING TO GROINS. 



PLATE 26. 

The Side Rib A, and the Angles being given straight upon the Plan, to find the Angle Rib G, 

and the Body Rib C. 

Let the rib A be supposed to be placed over the straight line a b, as its base, which 
divide into any number of equal parts, as eight, from the points of division draw lines to the 
centre of the groin to intersect the angles at a, b, c, d, e, f, g, these points will give the per- 
pendiculars of the ordinates of G, which, being made respectively equal to those of A, will 
give the curve of the rib G. If from the points a, b, c, &c, arcs be drawn from the centre 
of the groin to intersect the base of C, at 1, 2, 3, 4, 3, 2, 1, and perpendiculars be drawn 
and made correspondingly equal to those of A, and C be traced through these points, then 
C will be the body rib. 

How to describe the Ribs of a Groin over Stairs upon a circular Plan, the body Rib being 

given. 

Fig. 2. Take the tread of as many steps as you please, suppose nine, from E, and the 
heights corresponding to them, which lay down at F ; draw the plan of the angles as in the 
other groins, and take the stretch round the middle of the steps at E, and lay it from a to b 
at F ; make d e perpendicular to d c at B, equal to d e at F, draw the hypothenuse e c, draw 
perpendiculars from dcup to B, and prick B from A, as the figures direct, then B is the 
mould to stand over a b ; draw the chords a 4 and 4 m at the angles, make a 9, 4 h, perpen- 
dicular to them, each equal to half the height d e, at B or F, draw the hypothenuse g 4, 
and h m, draw the perpendicular ordinates from the chords through the intersection of the 
other lines that meet at the angles, then trace the moulds D and C, from the given rib A, 
will form the moulds for the angle or intersecting ribs. 

Note. The reason that the angle ribs D and C are laid contrary ways, is only to avoid confusion. « 



Plate 27. 




h c c . ,/ e 

//,/ If the ftacfcM&s 



CRADLING TO NICHES. 39 



PLATE 21. 

As all the sections of a sphere are circles, and those passing through its centre are equal, and the greatest which 
can be formed by cutting the sphere : it is therefore evident that if the head of a niche is intended to form a 
spherical surface, the most eligible method is to make the plane of the back ribs pass through the centre ; this 
may be done in an infinite variety of positions ; but perhaps the best, and that which would be easiest un- 
derstood, is to dispose them in vertical planes. If the head is a quarter of a sphere, the front rib, and the 
still plate or springing, on which the back ribs stand, will curve equally with the vertical ones ; but if other- 
wise, they will be portions of less circles. But it is evident if the front and springing ribs are intended 
to be arcs less than those of semicircles, either equal to each other or unequal, that as they are posited at 
right angles to each other, there can be only one sphere which can pass through them ; consequently if the 
places of the vertical ribs are marked on the plan, these ribs can have only one curve : in the former case 
no diagram is necessary, but in the latter it may be proper to show how the vertical ribs and their situation 
on the front rib are found. 

To get out the Ribs for the Head. 

From the centre C draw the ground-plan of the ribs as at figure A, and set out as many ribs upon the 
plan as you intend to have in the head of the niche,. and draw them all out towards the centre at C. Place 
the foot of your compass in the centre C, and from the ends of each rib, at e and c, draw the small con- 
centric dotted circles round. to the centre rib at m and n ; and draw m g and n i, parallel to r k, the face of 
the wall ; then from q round to c upon the plan, is the length and sweep of the centre rib, to stand over 
a b ; and from i round to e, the length and sweep of the rib that stands from c to d upon the plan ; and from 
g round to e is the sweep of the shortest rib, that stands from e to f upon the plan. 

Secondly. To bevel the Ends of the Back Ribs against the Front Rib. 

The back ribs are laid down distinct by themselves at C, D and E from the plan. Take c 1, in figure 
A, and set it from c to 1 in _D, will give the bevel of the top of the rib D. And from figure A, take from 
e to 2 upon the plan, and set from e to 2 in the rib E, will give the bevel of the top. 

TJiirdly. To find the places of the Back Ribs where they are fixed upon the Front. 

From the points a, c, and e, at the ends of the ribs, in the phn, figure A, draw the dotted lines up to 
the front rib, to df and w, which will show where they are to be fixed upon the front rib. The doable cir- 
cle upon the front rib shows the ranging. 



40 CRADLING TO NICHES. 



PLATE 28. 

To find the Curve of the Ribs of a Spherical Niche, the Plan and Elevation being given 

Segments of Circles. 

In fig. A is the elevation of the niche, being the segment of a circle whose centre is t; 
at B is the plan of the same width, and may be made to any depth, according to the place 
it is intended for, and its centre is c ; on the plan B, lay out as many ribs as it will require, 
draw them all tending to the centre at c, they will cut the plan of the front rib in g,f e, d; 
through the centre c, draw the line m n, parallel to a b, the plan of the front rib ; put the 
foot of your compass in the centre at c, draw the circular lines from a, g,f e, d, to the line 
m n, and make c s equal to u t, that is, make the distance from the middle of the chord line 
m n to s, the centre of the arch at C, equal to the distance from the middle of- the chord at 
the top at fig. A, to its centre at t; then place the foot of your compass in s, as a centre, and 
from the extremities m or n, describe the arch at C ; with the same centre draw another line 
parallel to it, to any breadth as you intend your ribs shall be ; then C is the true sweep of 
all the back ribs in the niche. 

Note. The points /, k, i, h, show what length of each rib will be sufficient from the point m ; from h to m is 
the rib that will stand over d x, from i to m is the rib that will stand over e w, from k to m overfv, and from I to m 
over g w : the other half is the same. 

Through the centre t, draw d e, parallel to a b, complete the semicircle e f g d, then 
d e is the diameter ; through n draw n a parallel to u d, in the centre t, with the dis- 
tance t a describe another semicircle whose diameter is c b; then will the semicircle 
c e g a b be equal to a vertical section of the globe, standing on k i, passing through its 
centre at c, which is the same curve as the rib at C, because u A is equal to c n, and c s 
bisecting m n at right angles, is equal to t u, bisecting e a at right angles : therefore the 
hypothenuse t a, that is, the radius of the circle b a g e c, is equal to s m or s n, the 
radius of the circle or rib at C. 



TLit* 2t 



Fid. -J- 



C r ~l* 




Tlate ■?(). 



Fia.A. 




CRADLING TO NICHES. 41 



PLATE 29. 

The Plan of a Niche in a circular Wall being given, to find the Front Rib. 

B is the plan given, which is a semicircle whose diameter is a b, and a, i, k, I, in, h, the 
front of the circular wall ; suppose the semicircle B, to be turned round its diameter a b, so 
that the point v may stand perpendicular over h in the front of the wall, the seat of the semi- 
circle standing in this position upon the plan will be an ellipsis ; therefore divide half the arch 
of B upon the plan into any number of equal parts, as five ; draw the perpendiculars 1 d, 
2 e, 3f, 4 g, 5 h, upon the centre c with the radius c h, describe the quadrant of a smaller 
circle, which divide into the same number of equal parts as are round B ; through the points 
1, 2, 3, 4, 5, draw parallel lines to a b, to intersect the others or the points d, e,f g, h, through 
these points draw a curve, it will be an ellipsis ; then take the stretch-out of the rib B, round 
1, 2, 3, 4, 5, and lay the divisions from the centre both ways at F, stretched out; take the 
same distances d i, e Jc,fl, g m, from the plan, and at F make d i, eh,fl, equal to them, 
which will give a mould to bend under the front rib, so that the edge of the front rib will be 
perpendicular to a, i, k, I, m. 

Note. The curve of the front rib is a semicircle, the same as the ground-plan, and the back ribs at C D and E 
are likewise of the same sweep. 

The reason of this is easily conceived, the niche being part of a globe, the curvature 
must be everywhere the same, and consequently the ribs must fit upon that curvature. 

Note. The curve of the mould F will not be exactly true, as the distances d i, e k,fl, &c, are rather too short 
for the same corresponding distances upon the soffit at F, but in practice it will be sufficiently near for plaster work ; 
but those who would wish to see a method more exact, may examine Plate 15, Jig. A, where C is the exact soffit that 
will bend over its plan at B. 

In applying the mould F when bent round the under edge of the front rib, the straight 
side of the mould F must be kept close to the back edge of the front rib, and the rib being 
drawn by the other edge of the mould, will give its place over the plan. 



42 CRADLING TO NICHES. 



PLATE 30. 

The Plan and Elevation of an Elliptic Niche being given, to find the Curve of the Ribs. 

Fig. A. Describe every rib with a trammel, by taking the extent of each base from the 
plan whereon the ribs stand to its centre, and the height of each rib to the height of the top 
of the niche, it will give the true sweep of each rib. 

To back the Ribs of the Niche. 

There will be no occasion for making any moulds for these ribs, but make the ribs them- 
selves ; then there will be two ribs of each kind ; take the small distances 1 e, 2 d, from the 
plan at B, and put it to the bottom of the ribs D and E, from d to 2, and e to 1 ; then the 
ranging may be drawn off by the other corresponding rib ; or with the trammel, as for 
example, at the rib E, by moving the centre of the trammel towards e, upon the line e c, 
from the centre c, equal to the distance 1 e, the trammel rod remaining the same as when the 
inside of the curve was struck. 

Given one of the common Ribs of the bracketing of a Cove, to find the Angle Bracket for a 

rectangular Room. Fig. F. 

Let H be the common bracket, b c its base ; draw b a perpendicular to b c, and equal to 
it draw the hypothenuse a c, which will be the place of the mitre ; take any number of ordi- 
nates in H, perpendicular to b c, its base, and continue them to meet the mitre line a c, 
that is, the base of the bracket at I; draw the ordinates of I" at right angles to its base ; 
then the bracket at J, being pricked from H, as may be seen by the figures, will be the 
form of the angle rib required. 

Note. The angle rib must be ranged either externally or internally, according to the angle of the room. 

Having given a common Bracket K, Fig. G,for Plaster Cornice, to find the Mitre Bracket L. 
Proceed as in the last example, and you will have the bracket required. 



Plate 30 



Fiq. A 




Flak. 32. 



Fin. A, 




FigB 




Fw'iC 




Fig D, 




CRADLING TO DOMES, &c. 43 



PLATE 31. 

OF PENDENTIVES AND INTERIOR DOMES WHEN PLACED OVER THE OPENINGS 

OF ROOMS. 



One of the Ribs of a Dome being given, and the Plan of the Opening of a Staircase which is square, and an 
octagon Curb at the Top for a Sky-light ; to find the Ribs and the springing Curve on each Side of the 
opening of the Staircase, where the Foot of the Ribs come, so that Part of the Dome shall be an octagon 
finish, agreeably to the Curb. 

Fig. A. Let B be the given rib ; take any number of perpendicular ordinates to its base at pleasure, 
from the points a, c, e, g, i, I, where they intersect its base, draw parallel lines to the sides of the curb } 
returning round each diagonal, if there is more than one, till it cut the base of the angle rib D ; at the 
points a, c, e, g, i, I, draw the ordinates of D, and prick it from B, will be the angle rib ; and at the points 
e, g, i, I, at C, upon the side of the opening of the staircase, draw the perpendicular ordinates, and prick 
C from B, agreeably to the letters ; then the curve C will be the true place for the foot of the ribs upon the 
side of the staircase, and the part that lies in the middle is a straight line parallel to the horizon. 

The Opening or Plan of the Room being a Square as before, and the vertical Section of a semicircular, to find 
the springing Curve D on the Side of the Room for the Foot of the Ribs, so that it shall finish to a cir- 
cular Curb at the Top. 

On the side of the staircase I m, as a diameter, describe a semicircle ; D will be the true place for the 
foot of the ribs ; this is evident, for every section of a semi-globe, at right angles to its base, is a semicir- 
cle, and this is the same thing if truly considered. 

Note. All the ribs of this dome are cut by the rib at C, as explained by the perpendicular lines ; draw round the 
centre a, from the points of each bracket, at c d ef, to the points k i h g, from these points draw perpendicular 
dotted lines, and these will show what length each bracket must have according to its place. 

The vertical Section of a Segment Dome passing through its Centre being given, the Plan of the Opening oj 
the Room being still a Square, as before, to find the Section upon each Wall for the Springing of the Ribs, 
to finish to a circular Curb at the Top. 

Let the section D across the angle be given, whose centre is k, and the distance of the centre from the 
chord k I ; bisect the side c g of the wall b h, at right angles at the point i ; from i make i b equal to I k ; 
with a radius b g or b c, describe the segment c, m g will be the true place of springing of the ribs ; all the 
other lesser ribs are cut from the angle rib D: all this is evident from the sections of a globe, which is 
already described in the Geometry. 

Fig. D is of the same nature as the others, having an ogee top ; the section F is traced from E. 



44 CRADLING TO DOMES. 



PLATE 32. 

Fig. A is the plan of an elliptical domical sky-light over a staircase ; B and C are the 
sections, which show how to place your ribs. 

How to proportionate the Length of the inside Curb to any Width given. 

Proceed as directed in page 48 for an elliptical dome, that will determine the true length 
to the width. 

How to proportionate the circumscribing Ellipsis, to pass through the Angles at a, b, c, and d, 
to have the same Proportion as a b, and be, o/ the Sides of the Staircase. 

Proceed as directed in fig. 5, Plate 6, in the Geometry. 

To describe the Ribs. 

The rib over from n, to the centre of the trammel in fig. A, is a given quarter of a cir- 
cle, as is shown at F, and of course all the other ribs must come to the same height with it. 
Suppose it was required to find a rib over dp, you must take the full extent from d to the 
centre, and describe the quarter of an ellipsis D; then the part over dp will be as much of 
it as is wanted : in the same manner E will be described, and the part over i o is what is 
wanted of this rib ; the same letters are marked upon the bases of D and E, as they are in 
the plan jfo?. A. Every other rib is described in the same manner. 

To find the Springings on each Side of the Room for the Foot of the Ribs to stand upon. 

Describe the semicircle C, to c b, the width of the room, and it will give the bottom of 
the ribs on that side ; and describe a quarter of the ellipsis B, for the bottom of the ribs on 
the other side, to the same height as C. 

This method depends on this principle, that all the parallel sections of a spheroid are 
similar figures : therefore a vertical section standing upon a b, will be similar to a vertical 
section passing through its centre ; both will be similar ellipses ; but a b is an ordinate to 
the conjugate axis, and b c is an ordinate to the transverse of the circumscribing ellipsis ; by 
construction half the length of the parallelogram is to half the length of the ellipsis, as half 
the width of the parallelogram is to half the width of the ellipsis, and a spheroid may be 
supposed to be generated by the revolution of a semi-ellipsis about its axis ; hence it follows, 
that all sections of a spheroid parallel to the axis are similar figures, consequently the section 
B is similar to the circumscribing ellipsis of the ground plan. 



Plate J2. 






Plate 33. 




Width of the Roof 







Tie Beam ( 




\ I* ■ 




WaU Plate 





ROOFING. 45 



PLATE 33. 

Let a b be the end of a rectangular roof, a e and bf being a part of each side, let a e 
and bf be each equal to half the width a b of the roof; join ef; bisect ef in c; draw c d 
perpendicular to ef and make c d equal to the height of the roof; join d e and df and d e 
and dfare the length of the principal rafters; join a c and b c; produce either diagonal, as 
b c to g, make c g equal to c d; join a g, and a g is the length of each hip. 

Draw any line x y perpendicular to the seat a c of a hip, cutting a e and a b at a; and y, 
and a c at 2- : from z describe a tanged circle to a g, cutting acatro; join w x and w y, 
and the angle y w x is the inclination of the planes which form the hip angle, and is what is 
generally termed the backing of the hip. 

In this plate one end of the roof is shown in order to show two cases : the first is when 
the purline lies level, or having two sides parallel to the horizon; the square at B, and the 
bevel at C, will show how to draw the end of the purline in this easy case ; but the following 
method is universal in all positions of the purline. 

Note. There will be no occasion to draw this at large ; as the bevels will be the same if done to ever so small 
a scale, and the sides may be measured from the scale. 

To find the Bevels of a Purline against the Hip Rafter. 

Let the purline be in any place of the rafter, as at /, and in its most common position, 
that is, to stand square, or at right angles to the rafter ; and from the point h as a centre, 
with any radius describe a circle. Draw two lines q I and p n, to touch the circle in p and 
q, parallel Xofb; and at the points s and r, where the circle and two sides of the purline 
intersect, draw two parallel lines to the former, to cut the diagonal in m and k ; and draw 
m n and k I perpendicular to s m and r k, and join the points n, i and k, i ; then G is the 
down bevel, and F the side bevel of a purline : these two bevels, when applied to the end 
of the purline, and when cut by them, will exactly fit the side hip rafter. 

To find the Bevels of a Jack Rafter against the Hip. 

By turning the stock of the side bevel of the purline, at F, from a round to the line i z, 
will give the side bevel of the jack rafter. And the bevel at A, that is, the top of a com- 
mon rafter, is the down bevel of the jack rafter. 

At the bottom is shown the manner of cocking down the tie beam upon the wall 
plate ; the proper size of the cocking is figured. 



46 ROOFING. 



PLATE 34. 

This plate shows the manner of framing a roof in ledgement ; but as roofs are seldom 
executed in this manner, I shall not be very particular in describing its lines. The fol- 
lowing description for winding will serve for any. 



Plate 3 /. 




Flute 35. 




ROOFING. 47 



PLATE 35. 

How to lay out an irregular Roof in Ledgement, with all its Beams lying bevel upon the Plan, 
so that the Ridge may be level when finished ; the Plan and Height of the Room being given. 

The lengths of the common and hip rafters are found as usual. From each side in the 
broadest end of the roof, through c and d, draw two parallel lines to the ridge line; draw 
lines from the centres and ends of the beams perpendicular to the ridge line, and lay out the 
two sides of the roof D and E, by making e d at E, equal to x n in A, the length of the long- 
est common rafter and c a in E, equal to u v at A, and so on with all the other rafters. 

To find the Winding of this Roof. 

Take y v half the base of the shortest rafter : and apply this to the base of the longest 
rafter from z to 1 ; then the distance from 1 to 2 shows the quantity of winding. 

How to lay the Sides in Winding. 

Lay a straight beam along the top ends of the rafters at E, that is, from c to e, and lay 
another beam along the line a b, parallel to it, to take the ends of the hip rafters of m and /, 
and the beams to be made out of winding at first. Raise the beam that lies from a to b, at 
the point b, to the distance 1 2 above the level; which beam, being thus raised, will raise 
all the ends of the rafters gradually, the same as they would be when in their places. 

The same is to be understood of the other side D ; the ends are laid down in the same 
manner as making a triangle of any three dimensions. 

To satisfy the curious, I have given the lines of this roof; but in practice there is not 
the least occasion for framing the sides in winding ; for, instead of the ridge line, the top is 
made level at the widest end of the roof, from the narrowest end, which begins at a point ; 
and by this means the sides may be framed quite out of winding, which will have a much 
better effect than any winding roof can have. 



48 ROOFING. 



PLATE 36. 

POLYGON ROOFS. 

The methods of constructing regular polygons upon any given side, are shown at figs. 4, 5, and 6, in 
Geometry. 

The Plan of a Polygon Roof being given, and one of the common ribs standing upon that Plan, to find the 
Angle Rib, and the Form of the Boards that will cover it when the Ribs are fitted up. 

In fig. A let B be the given rib ; divide the curve in any number of equal parts, as four, and lay 
them at D from a to 4, which bisects b b, the side of the polygon, at right angles; through these points 
draw lines parallel to the side & & of the polygon ; at B and D make 1 c at D equal to c c, between B and 
C make 2 d equal to d d, and 3 e equal to e e, Sac, and through the points b, c, d, e,f draw a curve line, 
which will be the form of the boarding ; from the points g,f, e, d, c, draw lines at right angles to g b, the 
base of the angle rib, and prick the rib C from B, as they are marked by the letters, which is plain. 

Note. The more parts there are in this operation, the truer will it be, or any other of this nature. 

In the same manner may the covering and angle ribs of any other polygon be found, whatever may be 
the form of the ribs, as is shown at figures B and C. 

To find the Covering of a spherical Dome. 

Fig. D. Make a circle i c kf, of the plan of the dome, and if it is a semi-globe, take the stretch-out 
of one quarter for the length of a board ; make the length of K from a to 4 equal to it, and let c c, at the 
bottom, be any breadth that the board will admit of; on the base c c as a diameter, make a semircle; divide 
half the arch line into any number of equal parts; draw the little lines 1 1, 2 2, 3 3, parallel to c c, the 
base of the board, and divide the height into the same number of equal parts ; draw the ordinates across ; 
make 1 1, 2 2, 3 3, upon these ordinates, equal to 1 1, 2 2, 3 3, in the semicircle at the bottom ; a curve 
being drawn through these points will be the mould K for the covering. 

To cover a spherical Dome when the Top does not rise so high as a Semicircle, but only a Segment. 

Suppose Id to be the height of the dome at F, and the width cf of the dome as before, upon the 
chord cf, with the perpendicular height I d describe a segment, which will be the same as a vertical sec- 
tion standing upon cf; here is only one. half of the segment, which is sufficient : draw the chord c d ; take 
c a equal to half the width of a board, whatever it will admit of: draw a b perpendicular to cut the chord 
c d at b ; take the stretch or circumference of the arch c d, and make the length of /from a to 4 equal to 
it ; take the double of a c, at F, and make it the base of the board at I; take a b from F, and set it upon 
the base of I, upon the middle of c c from a to b ; and with the chord c c, and the height a b, describe a 
segment upon the bottom of the board at J; divide one-half into any number of equal parts; likewise divide 
the height of the board J into the same number of equal parts; draw ordinates in both, and the board 1 
will be completed, as in the same manner as that of H, described before. 



FhiteJti 




Pla/r 37. 





ROOFING. 49 



PLATE 37. 

CIRCULAR DOMES. 

As the common method of finding the centres for describing the boards to cover a horizontal dome, 
will be found in practice very inconvenient, for those boards which come near to the bottom, I shall in this 
place show how to remedy that inconvenience. 

To find the Sweep of the Boards on the Top. Fig. A. 
Divide round the circumference of the dome into equal parts at 1, 2, 3, 4, 5, 6, &c, each division to 
the width of a board, making proper allowance for the camber of each board ; draw a line through the points 
1, 2, to meet the axis of the dome at x ; on x, as a centre, with the radii x 1 and x 2 describe the two 
concentric circles, it will form the board G ; in the same manner continue a line through the points 2 and 
3 at C, to meet the axis in w ; then w is the centre for the board C ; proceed in the same manner for the 
boards D, E, and F. 

Now suppose F to be the last board that you can conveniently find a centre, for want of room ; on t 
its centre, and the radius t 5, make from t on the axis of the dome t a, equal to t 5; through the points 5 
and a draw the dotted line 5 a b, to cut the other side of the circumference of the dome at b ; from the 
points 6, 7, 8, 9, 10, 11, draw radial lines to 5, to cut the axis of the dome at i, k, I, m,n,o; also through 
the points 6, 7, 8, 9, 10, 11, draw the parallels 6 c, 7 d, 8 e, &c, then will each of these parallel lines be 
half the length of a chord line for each board; then take c 6 from fig. A, which transfer to No. 1, from c to 
6 and 6 ; make the height c i, at No. 1, equal to c i, at fig. A ; and draw the chords i 6 and i 6 ; then upon 
either point 6, as a centre with any radius, describe an arch of a circle 12; divide it into two equal parts 
at 1, and through the points 6 and 1 draw 6 q ; bisect i 6, in^> ; draw p q perpendicular; then i 6 is the 
length, and p q the height of the board G, which may be described as in fig. 4, plate 5, of the Geometry. 
The reader must observe, that the length of a board is of no consequence so as the true sweep is got, which 
is all that is required. Proceed in the same manner with No. 2, by taking d 7 frora^g-. A, and place it at 
No. 2, on each side of d at 7 and 7, and take d k, horn fig. A, and make d k at No. 2, equal to it ; draw 
the chords k 7 and k 7, and bisect k 7 at n; draw n a perpendicular; upon the other extremity at 7, as a 
centre, describe an arch 12, and bisect it at 1, and through the points 7 and 1 draw the line 7 a, to cut 
the perpendicular n a at a ; but if the distance A; 7 is too long for the length of a board, bisect the arch 1 
at b; through 7 and b draw 7 t, and draw the little chord a 7, and bisect it at t ; draw t u perpendicular to 
intersect 7 4 at u ; and with the chord 7 a and the height t u, describe the segment H. 

In the same manner may the next board i" be found, and by this means you may bring the sweep of 
your board into the smallest compass, without having any recourse to the centre. 

Suppose it were required to draw a Tangent from 8 at No. 3, without having recourse to the Centre. 
Bisect the arch 8 I 8 at I ; on 8 as a centre, with a radius 8 I, describe an arch e I t ; make 1 1 equal 
to I e ; draw the tangent t 8. 

Given three Points in the Circumference of a Circle, to find any Number of equidistant Points beyond those 

that will be in the same Circumference. 
Fig. K. Suppose the three points «, b, c, to be given ; to one of the extreme points a join the other 
two points b and c by the lines a b and a c ; with a radius a b, and the centre a, describe the arch of a circle 
6 12 3; then take 6 1, and set it from 1 to 2, and from 2 to 3 ; through the points 2 and 3, draw a d 
and ae; then take b c, put the foot of your compass in c, and with the other foot cross the line a d at d; 
with the same extent put the foot of your compass in d, and with the other foot cross the line a e at e ; in 
the same manner you may proceed for any number of points whatever. 
7 



50 ROOFING. 



PLATE 38. 

SPHEROIDAL DOMES. 

Fig. A is the plan of a spheroidal dome ; B is the longest section, C the shortest section ; at a a in B, 
and b b in C, shows how to square the purlines, so that one side may be fair with the surface of the dome ; 
the dotted lines from a a in B, and b b in C, show how to get the length and width of the purline in fig. 
A ; but if the sides of the purline were made to stand perpendicular over the plan, the curve of it would 
be found in the same manner as before; then it would require no more than half the stuff that the other 
would, and take only half the time in doing, which is a considerable advantage. 

How to proportionate the inside Curve for the Sky-light, so as it shall answer to the Surface of the Borne. 

Draw the diagonal i l and k m in fig. A, and let h e or gf be the width, then h g or ef will be the true 
length of the curb ; because every section parallel to the base will be proportional to the base. 

To find the Ribs for this Dome. 

The ribs in this are got in the same manner as the ribs for a niche, as directed in page 41 ; and if the 
reader understand that, he must know this. 

To find the Form of a Board to stand in any Place of the Dome, in order to be bent up to the Crown. 

Divide one quarter of the base of the dome at D into three equal parts a b, b c, c d, and suppose you 
would find a board over a b c in the plan ; draw a c, b c, c c, and d c, to the centre at c ; then take the 
triangle a b c in D, and lay it down at a b c in G ; then draw the line c 1 1 1, &c, at right angles to a 6, 
and describe a rib G to the height of the dome, and the length to the perpendicular of the triangle a b c, 
and divide it into five equal parts, lay them along the line 111, &c, in H, and prick the mould H from 
the triangle a b c, as the letters are marked. The board K will be found in the same manner. 

Note. In the practice, you are to divide one-quarter of this dome into as many parts as you think the breadth 
of a board will contain ; and the boards, when got out by this method, will fit to a very great exactness; this is only 
into three, that the parts may be clearly seen to learners. 

If the boards are got out for one quarter to the lines here laid down, the boards that are in the other 
three quarters will not require any other lines, for every board in the first quarter will be a mould for three 
more boards. 



/V,//v 33 




i a i 




INTRODUCTION 



TO 



PRACTICAL CARPENTRY. 



OF THE COMPARATIVE STRENGTH OF TIMBER. 



PROPOSITION I. 

THE strengths of the different pieces of timber, each of the same length and thickness, 
are in proportion to the square of the depth ; but if the thickness and depth are both to be 
considered, then the strength will be in proportion to the square of the depth, multiplied into 
the thickness ; and if all the three dimensions are taken jointly, then the weights that will 
break each will be in proportion to the square of the depth multiplied into the thickness, and 
divided by the length ; this is proved by the doctrine of mechanics. Hence a true rule will 
appear for proportioning the strength of timbers to one another. 

RULE. 

Multiply the square of the depth of each piece of timber into the thickness ; and each product 
being divided by the respective lengths, will give the proportional strength of each. 

EXAMPLE. 

Suppose three pieces of timber, of the following dimensions : 
The first, 6 inches deep, 3 inches thick, and 12 feet long. 
The second, 5 inches deep, 4 inches thick, and 8 feet long. 

The third, 9 inches deep, 8 inches thick, and 15 feet long. The comparative weight 
that will break each piece is required. 



52 INTRODUCTION TO PRACTICAL CARPENTRY. 







OPERATIONS. 








First. 
6 deep 
6 




Second. 
5 deep 
5 






Third. 
9 deep 
9 


36 
3 thick 


Len 


25 
4 thick 


half 


Length 


81 
8 thick 


Length 12)108 


gth 8)100 


15)648(43 and a fifth 
fiO 


9 


12 and a 






48 
45 



Therefore the weights that will break each are nearly in proportion to the numbers 9, 12, 
and 43, leaving out the fractions, in which you will observe, that the number 43 is almost 5 
times the number 9 ; therefore the third piece of timber will bear almost 5 times as much 
weight as the first ; and the second piece nearly once and a third the weight of the first piece ; 
because the number 12 is once and a third greater than the number 9. 

The timber is supposed to be everywhere of the same texture ; otherwise these calcula- 
tions cannot hold true. 

PROPOSITION II. 

Given the length, breadth, and depth of a piece of timber ; to find the depth of another 
piece whose length and breadth are given, so that it shall bear the same weight as the first 
piece, or any number of times more. 

RULE. 

Multiply the square of the depth of the first piece into its breadth, and divide that pro- 
duct by its length : multiply the quotient by the number of times as you would have the 
other piece to carry more weight than the first, and multiply that by the length of the last 
piece, and divide it by its width ; out of this last quotient extract the square root, which is 
the depth required. 

EXAMPLE I. 

Suppose a piece of timber 12 feet long, 6 inches deep, 4 inches thick; another piece 20 
feet long, 5 inches thick ; requireth its depth, so that it shall bear twice the weight of the 
first piece. 



INTRODUCTION TO PRACTICAL CARPENTRY. 



53 



6 deep 
6 



36 

4 



12)144 



12 

2 times 



24 

20 length 



5)480 



Proof. 
9-7 
9-7 

67-9 
873 

94-09 
1-91 remainder added 



96-00 
5 



20)480 



24 



96)9-7 or 9-8, nearly for the depth 
81 



187)1500 
1309 



191 



EXAMPLE II. 



Suppose a piece of timber 14 feet long, 8 inches deep, 3 inches thick ; requireth the 
depth of another piece 18 feet long, 4 inches thick, so that the last piece shall bear five times 
as much weight as the first. 



64 
3 



As the length of both pieces of timber is divisible 
by the number 2, therefore half the length of each is 
used instead of the whole ; the answer will be the 
same. 



half 7)192 



27-4, &c. 
5 times 



137 

9 half the length 



4)1233 



308-25(17-5 the depth nearly 
1 



27)208( 
189 

345).1925, &c. 



54 INTRODUCTION TO PRACTICAL CARPENTRY. 



PROPOSITION III. 

Given thelength, breadth, and depth of a piece of timber ; to find the breadth of ano- 
ther piece whose length and depth are given, so that the last piece shall bear the same weight 
as the first piece or any number of times more. 

RULE. 

Multiply the square of the depth of the first piece into its thickness ; that divided by its 
length, multiply the quotient by the number of times as you would have the last piece bear 
more than the first ; that being multiplied by the length of the last piece, and divided by the 
square of its depth, this quotient will be the breadth required. 

EXAMPLE I. 

Given a piece of timber 12 feet long, 6 inches deep, 4 inches thick; and another piece 
16 feet long, 8 inches deep ; requireth the thickness, so that it shall bear twice as much 
weight as the first piece. 

6 
.6 



36 
4 



3)144 



48 
2 



96 

4 



8)384 
8)48 



6 thickness 



Or this at full length, 

6 depth of the first piece 
6 


36 
4 thickness of the first piece 


Length 12)144 


12 

2 by the number of times stronger 


24 

16 length of the last piece 


144 
24 


8)384 


8)48 


6 thickness 


EXAMPLE II. 



Given a piece of timber 12 feet long, 5 inches deep, 3 inches thick ; and another piece 
14 feet long, 6 inches deep ; requireth the thickness, so that the last piece may bear four 
times as much weight as the first piece. 



INTRODUCTION TO PRACTICAL CARPENTRY. 55 



25 
3 



12)75 



6-25 
4 



25-0 
14 



100 
25 



6)350 
6)58-333 
9-722 

PROPOSITION IV. 

If the stress does not lie in the middle of the timber, but nearer to one end than the 
other, the strength in the middle will be to the strength in any other part of the timber, 
as 1 divided by the square of half the length is to 1 divided by the rectangle of the two seg- 
ments, which are parted by the weight. 

EXAMPLE I. 

Suppose a piece of timber 20 feet long, the depth and width are immaterial ; suppose the 
stress or weight to lie five feet distant from one of its ends, consequently from the other end 

1 111 

15 feet, then the above proportion will be = : = — as the strength at 

10x10 100 5x15 75 

100 100 1 

five feet from the end is to the strength at the middle, or ten feet, or as = 1 : = 1 -. 

100 75 3 

Hence it appears that a piece of timber 20 feet long is one-third stronger at 5 feet distance 
from the bearing, than it is in the middle, which is 10 feet, when cut in the above proportion. 

EXAMPLE II. 

Suppose a piece of timber 30 feet long ; let the weight be applied 4 feet distant from one 
end, or more properly from the place where it takes its bearing, then from the other end it 



56 INTRODUCTION TO PRACTICAL CARPENTRY. 

1111 

will be 26 feet, and the middle is 15 feet ; then, = : == , or as 

15x15 225 4x26 104 
225 245 17 1 

= 1 : = 2 , or nearly 2 -. 

225 104 104 6 

Hence it appears that a piece of timber 30 feet long will bear double the weight, and 
one-sixth more, at four feet distance from one end, than it will do in the middle, which is 
15 feet distant. 

EXAMPLE III. 

Allowing that 266 pounds will break a beam 26 inches long, requireth the weight that 
will break the same beam when it lies at 5 inches from either end ; then the distance to the 
other end is 21 inches ; 21 x5 = 105, the half of 26 inches is 13 .-. 13 x 13 = 169; therefore 
the strength at the middle of the piece is to the strength at five inches from the end, as 
169 169 169 

:: or as 1 : the proportion is stated thus : 

169 105 105 

lb. 
169 

1 : :: 266 : to the weight required 

105 

169 



2394 
1596 
266 

105)44954(428 
420 



295 
210 

854 
840 

14 



From this calculation it appears, that rather more than 428 pounds will break the beam 
at 5 inches distance from one of its ends, if 266 pounds will break the same beam in the 
middle. 

By similar propositions the scantlings of any timber may be computed, so that they shall 
sustain any given weight ; for if the weight one piece will sustain be known, with its dimen- 
sions, the weight that another piece will sustain, of any given dimensions, may also be com- 
puted. The reader must observe, that although the foregoing rules are mathematically true, 



INTRODUCTION TO PRACTICAL CARPENTRY. 57 

yet it is impossible to account for knots, cross-grained wood, &c, such pieces being not so 
strong as those which are straight in the grain ; and if care is not taken in choosing the tim- 
ber for a building, so that the grain of the timbers run nearly equal to one another, all rules 
which can be laid down will be baffled, and consequently all rules for just proportion will be 
useless in respect to its strength. It will be impossible, however, to estimate the strength 
of timber fit for any building, or to have any true knowledge of its proportions, without some 
rule ; as without a rule everything must be done by mere conjecture. 

Timber is much weakened by its own weight, except it stand perpendicular, which will 
be shown in the following problems ; if a mortice is to be cut in the side of a piece of timber, 
it will be much less weakened when cut near the top, than it will be if cut at the bottom, 
provided the tenon is driven hard in to fill up the mortice. 

The bending of timber will be nearly in proportion to the weight that is laid on it; no 
beam ought to be trusted for any long time with above one-third or one-fourth part of the 
weight it will absolutely carry ; for experiment proves, that a far less weight will break a 
piece of timber when hung to it for any considerable time, than what is sufficient to break 
it when first applied. 

PROBLEM I. 

Having the length and weight of a beam that can just support a given weight, to find 
the length of another beam of the same scantling that shall just break with its own weight. 

Let / = the length of the first beam ; 
L = the length of the second ; 
a = the weight of the first beam ; 
w = the additional weight that will break it. 
And because the weights that will break beams of the same scantling are reciprocally as their 
lengths, 

a 

TV + - 

11 a 2 

therefore — : — :: w + -: / = W= the weight that will break the greater beam ; be- 

l L 2 L 

a 
cause w -\ — is the whole weight that will break the lesser beam. 
2 
But the weights of beams of the same scantling are to one another as their lengths : 
a : L a 

Whence, I : L :: - = W half the weight of the greater beam. 

2 21 
8 



58 INTRODUCTION TO PRACTICAL CARPENTRY. 

Now the beam cannot break by its own weight, unless the weight of the beam be equal 
to the weight that will break it : 

a 
w + - 
La 2 2w + a 

Wherefore, = 1 = I 

21 L 21 

L 2 a=2w + a x I 2 
.-. a : 2 w + a :: I 2 : L 2 , consequently </ L 2 =L= the length of the beam that can just 
sustain its own weight. 

PROBLEM II. 

Having the weight of a beam that can just support a given weight in the middle, to find 
the depth of another beam similar to the former, so that it shall just support its own weight. 
Let d = the depth of the first beam ; 
x == the depth of the second ; 
a = the weight of the first beam ; 
w = the additional weight that will break the first beam. 

a Qw+a 

then will w -\ — or = the whole weight that will break the lesser beam. 

2 2 

And because the weights that will break similar beams are as the squares of their lengths, 

2w+a 2x 2 x ivx 2 +a 

... d 2 : x 2 :: : = W 

2 2d 

the weights of similar beams are as the cubes of their corresponding sides : 

a ax 3 

Hence d 3 : x 3 :: - : = W 

2 2^3 

ax 2 x 2 ?v + x 2 a 



2d 3 2d 2 

a x — 2w + a x d 

.-.a:a + 2?v::d:x= the depth required. 
As the weight of the lesser beam is to the weight of the lesser beam together with the 
additional weight, so is the depth of the lesser beam to the depth of the greater beam. 

Note. Any other corresponding sides will answer the same purpose, for they are all proportioned to one ano- 
ther. 



INTRODUCTION TO PRACTICAL CARPENTRY. 59 

EXAMPLE. 

Suppose a beam whose weight is one pound, and its length 10 feet, to carry a weight of 
399.5 pounds, requireth the length of a beam similar to the former, of the same matter, so that 
it shall break with its own weight. 

here a = 1 
and w = 399.5 

then a + 2w;=800 = l+2x 399.5 
d= 10 
Then by the last problem it will be 
1 : 800 :: 10 
10 



8000 = x for the length of a beam that will break by its own weight. 

PROBLEM III. 

The weight and length of a piece of timber being given, and the additional weight that 
will break it, to find the length of piece of timber similar to the former, so that this last 
piece of timber shall be the strongest possible : 

Put / = the length of the piece given ; 
w = half its weight ; 
W = the weight that will break it ; 
x = the length required. 
Then, because the weights that will break similar pieces of timber are in proportion to the 
square of their lengths, 

W x 2 + w x 
.-. P : x 2 :: W + w : = the whole weight that breaks the beam; 

2 

and because the weights of similar beams are as the cubes of their lengths, or any other cor- 
responding sides, 

TV x 

then l 3 : x 3 :: w : the weight of the beam ; 

h 
Wx 2 + wx z w x 3 
consequently is the weight that breaks the beam = a maximum ; 

I 2 /3 

therefore its fluxion is nothing. 

3wx 2 x 

that is, 2 Wxx + 2w xx = nothing. 

I 
3 ?v x 

2 W+2w = 

I 
2 W= + 2 ?v 

therefore, x = I x 

3 w 



60 INTRODUCTION TO PRACTICAL CARPENTRY. 

Hence it appears from the foregoing problems, that large timber is weakened in a much 
greater proportion than small timber, even in similar pieces ; therefore a proper allowance 
must be made for the weight of the pieces, as I shall here show by an example. 

Suppose a beam 12 feet long, and a foot square, whose weight is three hundred weight, 
to be capable of supporting 20 hundred weight, what weight will a beam 20 feet long, 15 
inches deep, and 12 thick, be able to support? 

12 inches square 15 

12 15 



144 75 

12 15 



12)1728 225 

12 

144 



2.0)270.0 



135 



But the weights of both beams are as their solid contents : 

therefore 12 inches square 15 deep 

12 12 wide 



144 180 

144 inches = 12 feet long 240 length in inches 



576 7200 
576 360 
144 



43200 solid contents of the 2d beam 



20736 solid contents of the 1st beam 

20736:43200::3 144::135::21.5 by prop. 1. 

3 215 

cwt. lb. 

20736)129600(6 .. 28 = the weight of the 2d 67-5 
124416 beam 135 
270 



.5184 



112 12)2902-5 



10368 12)241-875 
5184 



5184 20-15625 
112 



20736)580608(28 



41472 31250 

15625 

165888 15625 

165888 



17-50000 
16 



30 
5 

8.0 



INTRODUCTION TO PRACTICAL CARPENTRY. 61 

21 cwt. 56 lbs. is the weight that will break the first beam, and 20 cwt. 17 lbs. 8 oz. the 
weight that will break the second beam ; deduct out of these half their own weight. 

20::17::8 
3::14::0 half 



17...3..8 



Now 20 cwt. is the additional weight that will break the first beam ; and 17 cwt. 3 lbs. 
8 oz. the weight that will break the second: in which the reader will observe, that 10 :: 3 :: 8 
has a much less proportion to 20, than 20 cwt. 17 lbs. 8 oz. has to 21 :: 56. From these ex- 
amples the reader may see that a proper allowance ought to be made for all horizontal beams ; 
that is, half the weights of beams ought to be deducted out of the whole weight that they will 
carry, and that will give the weight that each piece will bear. 

If several pieces of timber of the same scantling and length are applied one above 
another, and supported by props at each end, they will be no stronger than if they were laid 
side by side ; or this, which is the same thing, the pieces that are applied one above ano- 
ther are no stronger than one single piece whose width is the width of the several pieces col- 
lected into one, and its depth the depth of one of the pieces ; it is therefore useless to cut 
a piece of timber lengthways, and apply the pieces so cut one above another, for these pieces 
are not so strong as before, even if bolted. 

EXAMPLE. 

Suppose a girder 16 inches deep, 12 inches thick, the length is immaterial, and let the 
depth be cut lengthways in two equal pieces ; then will each piece be 8 inches deep, and 
12 inches thick. Now, according to the rule of proportioning timber, the square of 16 inches, 
that is, the depth before it was cut, is 256, and the square of 8 inches is 64 ; but twice 64 
is only 128, therefore it appears that the two pieces applied one above another is but half the 
strength of the solid piece, because 256 is double 128. 

If a girder be cut lengthways in a perpendicular direction, the ends turned contrary, 
and then bolted together, it will be but very little stronger than before it was cut ; for al- 
though the ends being turned give to the girder an equal strength throughout, yet wherever 
a bolt is, there it will be weaker, and it is very doubtful whether the girder will be any 
stronger for this process of sawing and bolting ; and I say this from experience. 

If there are two pieces of timber of an equal scantling (PI. 51, Fig. B), the one lying 

horizontal, and the other inclined, the horizontal piece being supported at the points e and 

f, and the inclined piece at c and d, perpendicularly over e and/*, according to the principles 

of mechanics, these pieces will be equally strong. But to reason a little on this matter, 

let it be considered, that although the inclined piece D is longer, yet the weight has less effect 



62 INTRODUCTION TO PRACTICAL CARPENTRY. 

upon it when placed in the middle, than the weight at h has upon the horizontal piece C, 
the weights being the same; it is therefore reasonable to conclude, that in these positions 
the one will bear equal to the other. 

The foregoing rules will be found of excellent use when timber is wanted to support a 
great weight ; for, by knowing the superincumbent weight, the strength may be computed to 
a great degree of exactness, so that it shall be able to support the weight required. The 
consequence is as bad when there is too much timber, as when there is too little, for nothing 
is more requisite than a just proportion throughout the whole building, so that the strength 
of every part shall always be in proportion to the stress ; for when there is more strength 
given to some pieces than others, it encumbers the building, and consequently the founda- 
tions are less capable of supporting the superstructure. 

No judicious person, who has the care of constructing buildings, should rely on tables 
of scantlings, such as are commonly in books; for example, in story posts the scantlings, 
according to several authors, are as follows : 

For 9 feet high 6 inches square. 

12 8 

15 10 

18 12 

Now, according to this table, the scantlings are increased in proportion to the height ; 
but there is no propriety in this, for each of these will bear weight in proportion to the num- 
bers 9, 16, 25, and 36, that is, in proportion to the square of their heights, 36 being 4 times 
9; therefore the piece that is 18 feet long, will bear four times as much weight as that piece 
which is nine feet long ; but the 9 feet piece may have a much greater weight to carry than 
an 18 feet piece, suppose double : in this case it must be near 12 inches square instead of 6. 
The same is also to be observed in breast-summers, and in floors where they are wanted to 
support a great weight; but in common buildings, where there are only customary weights 
to support, the common tables for floors will be near enough for practice. 

To conclude the subject ; it may be proper to notice the following observations which 
several authors have judiciously made, viz. that in all timber there is moisture, wherefore 
all bearing timber ought to have a moderate camber, or roundness on the upper side, for till 
that moisture is dried out the timber will swag with its own weight. 

But then observe, that it is best to truss girders when they are fresh sawn out, for by 
their drying and shrinking, the trusses become more and more tight. 

That all beams or ties be cut, or in framing forced to a roundness, such as an inch in 
twenty feet in length, and that principal rafters also be cut or forced in framing, as before ; 
because all joists, though ever so well framed, by the shrinking of the timber and weight of 



INTRODUCTION TO PRACTICAL CARPENTRY. 63 

the covering, will swag, sometimes so much as not only to be visible, but to offend the eye : 
by this precaution the truss will always appear well. 

Likewise observe, that all case bays either in floors or roofs do not exceed twelve feet 
if possible, that is, do not let your joints in floors exceed twelve feet, nor your purlines in 
roofs, &c, but rather let their bearing be eight, nine, or ten feet : this should be regarded in 
forming the plan. 

Also in bridging floors, do not place your binding or strong joists above three, four, or 
five feet apart, and that your bridging or common joists are not above ten or twelve inches 
apart, that is, between one joist and another. 

Also, in fitting down tie beams upon the wall plates, never make your cocking too large, 
nor yet too near the outside of the wall plate, for the grain of the wood being cut across in the 
tie beam, the piece that remains upon its end will be apt to split off, but keeping it near the 
inside will tend to secure it. See Plate 33, at the bottom, where the dimensions are figured. 

Likewise observe, never to make double tenons for bearing uses, such as binding joists, 
common joists, or purlines; for, in the first place, it very much weakens whatever you 
frame it into, and in the second place it is a rarity to have a draught to both tenons, that 
is, to draw both joints close ; for the pin in passing through both tenons, if there is a draught 
in each, will bend so much, that unless it be as tough as wire, it must needs break in 
driving, and consequently do more hurt than good. 

Roofs will be much stronger if the purlines are notched above the principal rafters, than 
if they are framed into the side of the principals ; for by this means, when any weight is 
applied in the middle of the purline, it cannot bend, being confined by the other rafters ; and 
if it do, the sides of the other rafters must needs bend along with it, consequently it has the 
strength of all the other rafters sideways added to it. 



DESIGNS OF ROOFS. 



PLATE 39. 

A and B show the method of trussing girders as is used by the greatest masters at this 
time. 

C is a horizontal section of B. 

D is a section of the butment, by cutting across a b in A. 

E and jP show the two sides of the king-bolt, at c in A, which is made with a wedge- 
way upon the top, so that it may force out the trusses upon the butments. 

The cores of the trusses ought not to be let close into the grooves of the side beams, but 
should be well secured at the ends and in the middle ; for suppose a weight to be laid upon 
the girder sufficient to bend it, if the braces are tightly fitted they will bend along with the 
beams, and consequently be of no use ; but if they are not fitted close, the girder can never 
have a curvature to any sensible degree, for the side beams cannot bend without shortening 
their length, but in the act of bending, the braces force upon the ends, and consequently act in 
opposition to the side beam. 

In tightening the truss work the head of your king-bolt ought to be greased, so that it 
may slide freely past the ends of the trusses ; proceed to turn the nut of the king-bolt, and 
another person to hit the head at c, with a mallet, which will make it start every time it is 
hit, and give fresh ease at every turning of the nut, so that you may camber the girder to 
any degree that you shall have occasion for, but generally not above an inch in twenty feet. 

Note. The sections B, E, and F, are to one-eighth part of the real size. 



Plate -3Q. 




, 




Plate iO. 






DESIGNS OF ROOFS. 65 



PLATE 40. 

This contains the most simple construction of roofs. 

Fig. A is calculated for a small building ; at one end of the collar beam is the Carpen- 
ter's boast, what they term a dove-tail tenon ; but I think rather a rule-joint, as it is worked 
out to a centre. This roof will do for an extent of 20 or 25 feet. 

Fig. B is a truss for a roof, the purlines to be notched upon the principal rafters, as will 
be described in the following plates of ledgment roofs ; this roof may be well calculated for 
an extent of 30 or 35 feet, the height one-fourth of the span for slate covering. 

Fig. C is a simple construction of a roof, for the segment finish of a dome, which will 
answer to any of the above extents. 



66 DESIGNS OF ROOFS, 



PLATE 41. 

Fig. A is a roof for the purlines to be framed in, and the common rafters to come fair 
with the principals. 

Fig. B is a roof calculated for a greater extent than any of the foregoing roofs, and may 
well extend 50 or 60 feet. Here likewise is shown the connection of the roof in the walls. 

Fig. C is a roof supported by two queen posts, instead of a king post, to give room for 
a passage or any other conveniency in the roof. 



Plate 41. 






PUUe 12. 




"^ 



Fig. E. 



DESIGNS OF ROOFS. 67 



PLATE 42. 

This roof is calculated for a span of seventy or eighty feet. 

You will observe in this, and the foregoing roofs, that the trusses are the same in num- 
ber as the purlines which they have to support ; for how absurd it is to give a roof more 
strength than necessary ! but, on the other hand, the consequences will be dangerous if 
too weak. 

Fig. A is a design of a roof for a theatre, which may extend from 80 to 90 feet. 

As it happens frequently in building that walls run across the roof, in such cases there 
will be little occasion for trussing the roof; then the purlines may be trussed, which will 
save one or two pair of principals, which is a considerable advantage. 

Fig. B is a roof of this kind, which shows the ends of the purlines, and C shows how 
to truss the purline. 

D, E and F are the methods of scarfing timber. 



68 DESIGNS OF ROOFS. 



PLATE 43. 

Fig. A. A design of a roof to finish with the parapet, when covered in the extent 
from 30 to 40 feet. 

Fig. B. A roof for a theatre, the extent from 70 to 80 feet. 

Fig. C. A design for a partition. 



: , ' (v Fini&h With, the fOtl'tmri ir/ie/l ivnt'nt iff ///, ,.i/,/// /.<;/// 

:>r to /< /;,/ . 




I design for a Fiuti fioit 




J'hilr 44. 




/■'/,/ A '. 



n=a 



n 



4 



DESIGNS OF ROOFS. 69 



PLATE 44. 

Fig. A is a curb roof, with a door in the middle of the partition ; the beam a b to run 
quite across the roof. 

Fig. B is a roof calculated for two rooms. 

Fig. C shows the method of framing a bridge floor. 



70 DESIGNS OF ROOFS. 



PLATE 45. 

Fig. A is the design of an M roof, which, is useful in some cases where the span is great, 
and no wall between, and the roof is required not to appear of a great height ; but this sel- 
dom happens in practice, for if there is any wall between the external walls, the roofs are in 
general made double, as is shown at figures B, C, and D. 



/'/a/* /.■;. 




Plate 46. 




DESIGNS OF ROOFS. 71 



PLATE 46. 

Fig. A is a design for a church, roof, the extent marked on the plate. 

Fig. B is a design of the same kind, but may be applied to an extent much greater. 
These two roofs, when finished, will be the same in every respect with the cylindro-cylindric 
arches described in Plate 23 ; as the manner there shown of fixing the ribs will not be differ- 
ent in this, I refer the reader to the description of that plate. 

Fig. C is another design for a church roof, where the ceiling over the galleries is to finish 
level. 



72 DESIGNS OF ROOFS. 



PLATE 47. 

Fig. A is a design for a domical roof; B shows the manner of framing the curb for it to 
stand upon, the section of the curb being also shown upon the bottom of Jig. A. 

N. B. This design is nearly the same as that constructed for the dome of the Pantheon, 
London, which was burnt down. 



Plate 17. 




Pica*. 4$. 




DESIGNS OF ROOFS. 73 



PLATE 48. 

Fig. A is another design for a domical roof; the bottom of it is made into a very narrow 
compass, in order to gain room within the dome. 

Figures B and C are designs for circular, and elliptical trusses for bridges, &c. These 
trusses may also be applied to roofs where there is no cavity wanted within. 



10 



74 DESIGNS OF ROOFS. 



PLATE 49. 

Fig. A is a design for a story post and breast-summers. 

Fig. B is a design for a bridge. C is a section across. D is part of the plan, which also 
shows the manner of fixing the piles. E shows half the plans of the bridgings. 



Plate /!). 




U 




~7T 



Flu A 








n 


n n 


n 


n n 






'II II II II II II ||; 


r -ll II II II II II II' 


L_ 












- II 






J 



. Fl/tte 50. 




^1 M~ 



~XI \r 



N, K K ' 



HAND-RAILING. 



PREFACE. 

In that elegant branch of the building art called Joinery, Stairs and Hand-railing take 
precedence. For the manner of finding the face and falling moulds, I have laid down cor- 
rect methods founded on the most obvious principles, and which have been put in practice 
by myself, and by those who attended the instructions given by me, in this art, some years 
since, and found to answer well in every case. * 

The superior advantages in every respect of the new plates on hand-railing, over those 
published in the former editions of this work, will, I trust, be deemed a sufficient reason for 
the change made by me in this department of the Carpenter's Guide. I have retained those 
plates on hand-railing, by P. Nicholson, which are considered useful, and hope that the 
alterations made in this department of his work, will meet the approbation of Carpenters gene- 
rally. In conclusion, I think it proper to say, that, for the method of finding the butt joints, 
Plate 55, I am indebted to an eminent stair-builder of this city, whose mechanical skill in 
joinery, I, with others, hold in high estimation. 

WILLIAM JOHNSTON, 

Architect, Philadelphia. 

PLATE 50. 

Plan and elevation of a newel post stair-case. Scale i an inch to a foot. 

To draw the Ramp. 

Make A B equal to A C, draw CD at right angles with the rail, also produce the 
horizontal line E until it intersects at D, which is the centre of the ramp. 



76 HAND-RAILING. 



PLATE 51. 

Plan of stair-case on Plate 50, to a scale of 3 inches to a foot. 



Plate 51 




hLJ 



i\ y 





Plate 52 . 




HAND- RAILING. 77 



PLATE 52. 

Plan of a stair-case, showing how to arrange the steps at the circle, as to allow the string- 
board to be formed, without the easing (usually made) on its lower edge, caused by making 
the tread of the steps greater or less at the circle, than at the flyers ; this plan. also admits of 
the balusters being the same length, and nearly the same distance apart as at the flyers. For 
manner of drawing, see Plate 53. 



78 HAND-RAILING. 



PLATE 53. 

Well Hole of Plate 52 to a larger scale. — Describe the equilateral triangle ABC, draw 
the right line E D, touching the face of the string board, produce C A and C B, to inter- 
sect it at D and E, then is the right line E D equal to the semi-circumference of the circle. 

Upon this line lay off the tread of a step O P, and draw lines from those points to 
the centre C, which gives the position of the risers K K, on the circular part of the string- 
board, at the points S S. Also the position of the same riser on the carriage is found, by 
adding to the distance D P or O E, whatever may be required to make a full tread, and 
place it from A to W, which gives the position of the risers N N, and K K on the car- 
riages. 



r/<i/t- 53. 



Trimina 




Plate 54. 



S/retitc/wut ot' outside falling mould,. 




HAND-RAILING. 79 



PLATE 54. 

Being a plan of the rail for Plate 53 ; the stretchout is found by the same method as in 
the foregoing plate. 



80 HAND-RAILING. 



PLATE 55. 

To draw the falling moulds for the rail piece as shown in Plate 56. Let fig. 1 be the 
pitchboard, and A the beginning of the circular part of rail, from the line A, place the 
stretchout of the inside and outside of the rail B B, and from the line C set up the height 
B F or B E, which is equal to a rise and a half. — Having found the stretchout and height, 
place half the thickness of the rail on each side of the central points F E and H, as denoted 
by the small circles, connect the upper heights F E with the lower one fZ, which completes 
the falling moulds. 

To cut the centre or butt joints, divide the distance F E into two equal parts, draw the 
line H L, and at right angles with this line draw the lines F J and E K, which gives the 
joint required. 



Flate 55. 




Plate 56. 




HAND-RAILING, 31 



PLATE 56. 

To draw the face mould for plan of stairs, Plate 53. Let fig. 1 be the plan of the rail, 
draw the chord line A, also the base line P, at any convenient distance above it; make the 
length of this line equal to the dotted line on the plan, and set the height N E of the fall- 
ing moulds, Plate 55, from to R draw R P, then draw ordinates through the plan perpen- 
dicular to the base line, touching the chord R P; also draw the ordinates B B, B B, &c, at 
right angles with this chord, and make the distances 1, 2, 3, 4, &c, on the plan equal to 
1, 2, 3, 4, &c, of the face moulds. 

The circular part being found, draw the straight part by keeping it parallel to the line 
of the joint S. — In applying this mould to the plank, care is to be taken that the chord R P 
be kept parallel to its edge. 

As this method of finding the face mould does not admit (if 3 inch plank be used) of 
more than 4 or 5 inches of straight rail being attached to the circular part, another method 
is shown on Plate 58, by which the straight rail can be extended at pleasure. 
11 



82 HAND-RAILING. 



PLATE 57. 

Are the same face moulds as shown on Plate 55, — and is here reproduced to show 
the method of finding the spring of the plank, before drawing the face mould, Plate 58. 

Let fig. 1 be the plan of the rail, draw the line A B, at right angles with C D, also 
draw the dotted line D B, which is a continuation of the greater side of the pitchboard, to 
intersect it at B— then is the distance O P the spring of the plank. 



P/,l/r 57. 




PleUe 58. 




HAND-RAILING. 83 



PLATE 58. 

To draw the face mould O. 

Let fig. 1 be the plan of the rail, and fig. 2, the pitchboard ; draw the ordinates 1, 2, 3, 4, 
&c, at right angles to A B; place the spring O P, Plate 57, from C to D ; draw D E, at right 
angles to A C. Make D E, fig. 2, equal to E D, fig. 1 ; draw E F, also A G parallel to it, 
draw the lines 0, 0, 0, &c, from the points made by the intersection of the ordinates with 
the convex and concave sides of the rail, and produce them to the line A G, — draw A H at 
right angles with C A, make i" J equal to E C, fig. 2, join / K, parallel to / K, draw the 
ordinates L L, &c, and with A as a centre, draw the concentric dotted lines to intersect with 
the line A H, and continue those lines parallel to A C and at the points of intersection with 
the ordinates, trace the face mould O. — The dotted line S shows the over wood for cutting 
the square or butt joint. 



84 HAND-RAILING. 



PLATE 59. 

Plan of a stair-case having 8 winders. 

The manner of finding the stretchout of the convex and concave sides of the rail is the 
same as in the foregoing plates. 

For the falling moulds and face moulds of this stairway, see Plates 60 and 61. 



Plate 59. 




Plate 60. 




HAND-RAILING. 85 



PLATE 60. 

To draw the falling moulds for stair-case, Plate 59. 

Let A B be the stretchout of the convex side of the rail, and A J, the straight portion 
attached to the circular part, — draw the flyers H and G, and the four winders CD E and F, 
also draw the base line B I, at any convenient distance below the point O ; raise the rail 5 
inches above the line of the noseings, from the point K to J, also make the distance A L equal 
to the distance A L, Plate 61, and proceed to draw the moulds as before described in Plates 
55 or 57. 



86 HAND-RAILING. 



PLATE 61. 

To draw the face moulds for the stair-case, Plate 59. 

Let fig. 1 be the plan of the rail, — draw the chord A B, also the line C D, parallel to it, 
make the lines E F G, equal in height to the lines B L I, Plate 60. Join G H, also draw 
I J parallel to C D, and at the point / let fall the perpendicular line / K. — Join K M, which 
gives the directing ordinate on the plan ; draw a sufficient number of ordinates to meet the 
chord N O, also draw I P at right angles with the line H G, and with the distance K M, 
from the point Jasa centre, bisect the line I P at P. Join J P, which gives the directing 
ordinate for the face mould — transfer the ordinates from the plan, and set them from the 
chord O N, and through the points trace the face mould. Take the distance i" S, set it from 
Y to Z on the chord, and join Z m ; with this bevel the edge of the plank must correspond 
before you apply the face mould to its upper and lower surface. 



PlaA 61. 




Plate 62. 




HAND-RAILING. 87 



PLATE 62. 

To draw the Scroll of a Hand-rail. 

Make a circle 3^ inches diameter, divide the diameter into three equal parts, — make 
the square in the centre equal one of these parts, and divide each of its sides into 6 equal 
divisions, with the centres 1, 2, 3, 4, &c, complete the outside revolution, set the width of the 
rail from A to B, and go the reverse way to complete the inside revolution. 

If a scroll of less diameter be required— draw the dotted lines C D for the straight part 
of the rail, and a scroll of one quarter revolution less is given. 



HAND-RAILING. 



PLATE 63. 

To draw the falling moulds for the scroll. 

Let fig. 1 be the pitch-board, and A B the stretchout of the convex side of the rail 
E F, Plate 62, — lay half the thickness of the rail on each side of the upper line of the pitch- 
board, shown by the small circle; draw the line D C, and the horizontal line F C, and com- 
plete the under edge of the mould by intersecting lines. 

In forming this easeing the mould should not come to a level at the joint, but continue 
to descend for a few inches past it. 



PlcUe 63. 




Plate 64. 




HAND-RAILING. 89 



PLATE 64 

Face mould for the twist of the scroll. B the pitchboard. 

Draw a sufficient number of ordinates through the plan, and from the points 1, 2, 3, &c, 
draw ordinates at right angles to the line A E, and transfer the distances 1, 7, 2, 8, &c, upon 
these lines, and trace the face mould through the points. 



12 



90 HAND-RAILING. 



PLATE 65. 

Plan of the curtail Riser. When the scroll is drawn, set the projection of the nosing 
without, and draw it equally distant from it, which will give the form of the curtail step, 
then set the distance which the riser is back from the line of nosings, and you will get the 
form of the curtail riser. 



Flan 65 




FlaLC 66. 




STAIR-CASING. 91 

PLATE 66. 

PART 1. 

To find the Moulds for executing a Rail with a Semicircle of Winders. 

Figure 1. The falling mould as here drawn does not follow the line of nosings, but is 
raised six inches, as is the practice with several handrailers, in order that the rail should not 
approach nearer to the nosings of the winders than to the flyers. This example is adapted 
to a stair with ten winders in the semi-circumference. 

Fig. 2. The plan and face mould of the lower quarter winders. 

Fig. 3. The plan and face mould of the upper quarter winders. ABC the convex 
side of the quadrantal part of the plan, C D a straight part intended to be wrought on the 
same piece with the twisted part. In this method, the plane of the top of the face mould is 
supposed to rest upon the upper extremities of three straight lines or slender rods perpen- 
dicular to the plane of the seat of the said face, and these three perpendicular lines to rise 
from three points, in a line dividing the breadth of the plan everywhere into two equal parts 
of the rail piece ; and these three points to be so situated that one may be at each extremity, 
one at the end of the quadrant, and one at the end of the straight piece, and the intermediate 
point at the intersection of a perpendicular drawn from the centre of the plan of the rail 
piece to a straight line joining the two extreme points. Let each of the upper extremities 
of the three perpendicular lines be called resting points, and let the feet of the perpendicu- 
lars be called the foot of the heights of the rail piece, which will therefore be the same as 
the seats of the resting points ; and let the three perpendicular lines themselves be called 
the heights of the rail piece, and their places distinguished by the lower height, the middle 
height, and the upper height. 

For fig. 2, let a be the foot of the upper height, b the foot of the middle height, and d 
the foot of the lower height; join a d; draw afib'g and d h perpendicular to a d, the middle 
one b g being drawn from the centre E. Let A, B, C, D, figure 1, be the points corre- 
sponding to A, B, C, D, in the convex side of the iplan, figure 2, and let AF,B G and D H 
be the heights of the rail ipiece, figure 1; make afi b' g, d h, figure 2, respectively equal to 
A F, B G, D H, figure 1 : }omfh, figure 2 : draw g i parallel to b d, cutting f h at i: draw 
i k parallel to g b, cutting b d at k : draw g I m perpendicular to f h, cutting f h at I; join 
b k ; from i, with the distance b k describe an arc at m ; join i m : draw D' P parallel to d a 
from the extremity D of the concave side ; produce the convex quadrant C B A to Q, and 
the concave quadrant C B' A to P, and radiate the line P Q from the centre E, which will 
complete the whole plan of the rail piece ; the part P A' A Q will make a sufficient allowance 
for the cutting of the joint. Draw ordinates parallel to b k cutting the chord D' P, the con- 
cave side of the plan, and the convex side of the same : produce b k to meet D' P in t, and 
produce m i to u, making i u equal to k t : through u draw v w parallel to/ h; from the 
points where the ordinates intersect D' P, draw lines parallel to afi b g, or d h, cutting v w. 
from the cutting points in v rv draw lines parallel to u m as ordinates : transfer the interior 



92 STAIR-CASING. 



PLATE 66. 

PART 2. 

ordinates from the plan to the face mould, also transfer the exterior ordinates of the plan to 
the face mould, applying them from the intersected points in the chord v n>, and through the 
points thus set off, trace the concave and convex curves, as also the straight part of the 
mould ; observe, however, that as the straight part of the mould is a parallelogram, if 
three points on two contiguous sides are found, joining the middle point to each of the 
other two, gives two sides ; each of the other two remaining sides is found by drawing a line 
parallel to its opposite side. 

In the same manner the mould figure 3, is to be found ; but the base line of the heights 
is taken upon any convenient line & R, parallel to A D,fig. 1, so as to shorten the height 
lines, as otherwise figure 3 would occupy more space than might be found at all times con- 
venient, and at any rate the shortening of the heights will shorten the time of drawing 
figure 3, as shorter lines can be drawn sooner than longer ones. The distance between the 
height lines of the upper rail piece is the same as those for the under rail piece. 

To find the spring of the plank fig. 4 : draw any straight line A B; from which cut off 
B d equal to g I, figure 2 : draw d C, figure 4, perpendicular to A B : make d C equal to 
b b', figure 2, and join B C, figure 4, then the angle A B C is denominated the spring of the 
plank, and the angle is said to be acute when the planes of the top and edge. of the plank 
form an acute angle with each other : but when these two form an obtuse angle with each 
other, the spring is said to be obtuse. 

To find the Spring of the Plane at an Obtuse Angle. . 

In figure 5, let A B be any straight line, which produce to d : make B d equal to B d, 
figure 3 : in figure 5 draw d C perpendicular to A B : made d C equal to a e, figure 3, then 
ABC will be the spring of the plank at the obtuse angle. 

Both these bevels are supposed to be applied from the top of the plank : but if the com- 
plemental angle of the obtuse spring bevel which answers to the upper wreath piece be 
taken, then the lower spring is applied from the top in order to give the spring of the lower 
wreath piece, and the complemental spring of the upper wreathed piece to the lower edge 
of the said piece. 

The reader will perceive that in figures 2 and 3, though the plan is the same in both, 
but their position inverted, the face mould of the lower wreath piece is much longer than that 
of the upper one. This circumstance is owing to the middle parts of the falling mould being 
raised over the nosings of the winders, and the more it is raised above the winders the greater 
will the face mould of the lower wreathed piece exceed the length of the mould of the upper 
wreathed piece, and unless that (if supposing a line passing through the middle of the falling 
mould bisecting the breadth of the same), the distance of the line thus passing be the same 
over the winders as over the flyers, the two face moulds can never be equal. 



Flab. 67. 



Ah Ji/t/n>r7 P/tt// Scsection of a staireaso. For its lines see tJu lu.rt Plate 




STAIR-CASING. 93 



PLATE 67. 

Plan and elevation of an elliptical stair-case. In every kind of stair-case whatever, the 
breadths of the heads of the steps are always reckoned on a line, bisecting their length, or 
at 18 inches distant from the rail. In this example, the steps are divided into equal parts, 
both at the rail and at the wall. This division will make the falling mould straight on the 
edges, and consequently will form an easy skirting, as well as an easy rail. 



94 STAIR-CASING. 



PLATE 68. 

How to draw the Face Moulds of an elliptic Stair. 

The plan and section being laid down as in Plate 67, the reader will observe, that the 
ends of the steps are equally divided at each end ; that is, they are equally divided round the 
elliptic wall, and also at the rail. In this plate, the rail is laid down to a larger size than 
that in the last plate : the plan of this rail must be divided round, into as many equal parts 
as there are steps ; then take the treads of as many steps as you please, suppose 8, and let 
h h at Jig. H, be the tread of 8 steps from H; on the perpendicular h m set up the height of 
as many steps, that is 8 ; and draw the hypothenuse m h, which will give the under edge 
of the falling mould. The reader will observe that this falling mould will be a straight line 
excepting a little turn at the landing and at the scroll, where the rail must have a little bend 
at these places, in order to bring it level to the landing and to the scroll ; then mark the plan 
of your rail in as many places as you would have pieces in your rail (in this plan are three); 
then draw a chord line for each piece to the joints ; also draw lines parallel to the chords, to 
touch the convex side of the plan of the rail ; from every joint draw perpendiculars to their 
respective chords. Now the tread of the middle piece at C being just 8 steps, the height 
of the section from h to m is 8 steps ; and the section m n is the same as m n on the falling 
mould, and the section h i is the same height as h i upon the falling mould ; draw a line to 
touch the sections, and complete your face mould as in the foregoing plates : in each of the 
other pieces at E and G, the number of treads being 6, therefore, from your falling mould 
set the stretch of six steps ; from h to h draw h /, parallel to h n, then u k I will give the 
height of the sections at D and E : everything else agreeably to the letters. 

Note. The stretch-out of 8 steps, or any other number, is not reckoned on the chord ; but it is the stretch-out 
round the convex side of the rail, or what most people call the inside. 




Plate 69. 





STAIR-CASING. 95 



PLATE 69. 

PART 1. 

To show the proper twist of a rail and thickness of stuff, also the face moulds for the same. 

Fig. 1. No. 1. A A' O' O be the plan of a rail : figure 2, the falling mould as com- 
pleted. Let the plan of the rail No. I, fig. 1, be so placed that the chord of A C be parallel 
to the base M of the falling mould fig. 2, and let B B' be the separation of the straight 
and circular parts of the rail, O' B' B being the quadrantal part of the rail, and B' A', 
A B the straight part : divide the concave side B C D O of the quadrantal part into equal 
parts at the points B, C, D, &c, O; through all the points B, C, D, &c, draw lines at right 
angles to the chord A O, cutting it in 1, 2, 3, &c, and produce them upwards to the points 
f, g, h, &c. ; let M B N, fig. 2, be the section of a step, upon M O make B A equal to B A, 
No. 1, figure 1 : extend the parts A B, B C, CD, &c. No. 1, figure 1, upon the base M O, 
figure 2, from A to B, from B to C, from C to D, &c. : from the points A, B, C, D, &c, draw 
lines perpendicular to M O, cutting the lower edge of the falling mould at E, F, G, H, &c, 
and the upper edge at I, J, K, L, &c. In figure 1, draw any line, a, b, c, d, &c, parallel to 
the chord A 0; make ae, bf, eg, dh, &c, respectively equal to A E, B F, C G, D H, &c, 
figure 2; also in figure 1, make a i, bj, c Jc, d I, &c, equal to A I, B J, C K, D L, &c. : 
through the points e,f, g, h, &c, figure 1, draw a curve; also through the points i,j, Jc, I, 
&c, draw another curve, and these two curves will complete the projection of the falling 
mould. From the point s, figure 1, radiate the lines CC, DD' &cc, cutting the convex side 
at C D', &c. : from the points A', B', C, D', &c, draw the lines A' i', B'f, C k', D' h,' &c. : 
draw e e', i i',jf, k k ', h h', &c, parallel to the chord A O : through the points *, /, k, &c., 
draw a curve until it intersect with the curve i,j, Jc, &c, this will form the projection of the 
top of the rail piece. In like manner the under parts which appear at q q', h h', will be com- 
pleted in the same manner, so that No. 2 is the whole appearance of the solid, q r' r q' and 
i i e e' being sections or the ends which join the contiguous parts of the rail. 



96 STAIR-CASING. 



PLATE 69. 

PART 2. 

To trace the face mould No. 3 ; join i' r, and let the perpendiculars cut i r at t, 1, 2, 3, 
&c, draw the ordinates 1 b u, 2 c v, 3 d w, &c, perpendicular to i r : make 1 b, 2 c, 3 d, &c, 
equal 1 B, 2 C, 3 D, &c, No. 1 : also in No. 3, 1 u, 2 y, 3 w, &c., equal to 1 U, also make 
x o equal to X O', i a! equal to / A', 2 V,3 W, and C : draw b y parallel to t a', and a y parallel 
to I b, and t a! y b will complete the straight part of the face mould : join r o; draw the con- 
cave curve b e d . . . . r, and the convex curve y u v w o, which completes the curve 

part and the whole of the face mould No. 3. 

Fig. 3 shows the projection and face mould for the quadrantal part of a rail; the princi- 
ples of projection and manner of drawing the face mould are the same as what have now 
been shown. This figure is introduced to show how much less wood the part of the rail 
requires from a quadrantal plan, than when a straight part of the rail is taken in, and the 
more of the straight rail that is taken in, the greater will be the deflection from the chord ; 
thus in figure. 3, the distance between the chord f r at any point to the nearest point of the 
projection, is much less than the distance i' r,fig. 1, from a corresponding point in i' r to the 
nearest point of the projection. 

The dotted lines, fig. 3, show another face mould, the chord being drawn from the one 
extreme point to the other, of the upper end of each section, which makes the face mould 
much shorter than it ought to be, and will therefore require a much greater thickness of 
stuff. From this false position^' r of the chord, some unskilful teachers of drawing direct 
their pupils to trace the face mould of a hand-rail. 



■ffafe 70. 






STAIR-CASING. 97 



PLATE 70. 

To find the Face Mould of a Stair-case, so that when set to its proper Rake, it will be perpen- 
dicular to the plan whereon it stands for a level Landing. 

In fig. A draw the central line b q, parallel to the sides of the rail ; on the right line b q 
apply the pitch-board of a flyer ; from q to a draw ordinates n d, m e, If, k g, and i k, at 
discretion, taking care that one of the lines, as k g, touch the inside of the rail at the point 
g, so that you may obtain the same point exactly in the face mould ; then take the parts q u, 
u v, v w, w x, x y, and apply them at B from q u, u v, v w, w x, x y, from these points draw 
the ordinates of B, and prick them from the plan, figure A 7 as the letters explain ; then B 
will be the mould required. 

To find the Falling Mould. 

Divide the radius of the circle fig. A, into four equal parts, and set three of these parts 
from 4 to b ; through n and v, the extremities of the diameter of the rail, draw b n and 
b v, to cut the tangent line at the points c and d; then will c d be the circumference of 
the rail, which is applied from c to d, at C, as a base line : make c e the height of a step ; 
draw the hypothenuse e d, at. the point e and d; apply the pitchboard of a common step 
at each end of their bases, parallel to c d, make df equal to d e, if it will admit of it, and 
by these lengths curve off the angles by the common method of intersecting lines ; then 
draw a line parallel to it, for the upper edge of the mould. 



13 



98 HAND-RAILING. 



PLATE 71. 

To draw a Falling Mould for a Rail having Winders all round the circular Part, thence to 

find the Face Mould. 

To describe every particular in this, would almost be repeating what has been already 
described ; the heights are marked the same upon the falling mould at jD, as they are at 
the face mould, which will give the heights of the sections of the rail ; and the face mould 
at C is traced from the plan B, according to the letters ; G shows the application of the 
mould to the plank; take the bevel at H, and apply it to the edge of the plank at D, 
and draw the line b c ; then apply your mould to the top of the plank, keeping one corner 
of it to the point b, and the other corner close to the same edge of the plank; then draw the 
top face of the plank by your mould; then take your mould, and apply it to the under side 
at c, in the same manner. 



//////- 7 J 




I'UUt 72. 




HAND-KAILING. 99 



PLATE 72. 

To draw the form of a Hand-rail 

hi fig. A make an equilateral triangle vrv t upon its width, and divide it into five equal 
parts, and from one part on each side draw z s and y w, then t, g and m are the centres, I m 
being made equal to / g ; the centres are found the same for the other side. 

The Form of a Rail being given, to draw the Mitre Cap. 

Let the projection of the cap be three inches and a half, and make the distance of 
the inside circle from the outside circle the projection of the nose on each side of the 
rail, and draw the mitre n o and p o ; then continue parallel lines down to the mitre p o ; 
put the foot of your compass in the centre of the cap, and circle the parallel lines round 
to a, c, e, g, and i, and draw the ordinates ab, c d, ef, &c, and then prick the cap to the 
rail according to the letters. 

How to draw the Form of the Cap for the Mitre to come to the Centre. 

It is only drawing the parallel lines from the rail to the mitre wherever it is, and 
circling them round to the ordinates, and so pricked from the rail, and the thing is done. 



100 DIMINISHING RULES FOR COLUMNS. 



PLATE 73. 

How to diminish the Shaft of a Column by the ancient Method. 

In. fig. A, describe a semicircle at the bottom : let a line be drawn through the diameter 
at the top, parallel with the axis of the column, till it intersects the semicircle at 1, at the 
bottom; then 1, 1 at the bottom will be equal to 1, 1 at the top; divide the arch into four 
equal parts, and through these points draw lines parallel to the base, the height of the column 
being also divided into the same number of parts, and lines drawn parallel to the base, then 
the column is to be traced from the semicircle, according to the figures. 

How to diminish the Column by Lines drawn from a Centre at a Distance, 

Fig. C. Take the diameter a b, at the bottom, set the foot of your compass in c at the 
top, and cross the axis in the point d, continue c d at the top, and a b at the bottom, to meet 
at e; then draw from e as many lines across the column as you please, and take the diame- 
ter a b at the bottom, and prick each line upon the axis equal to b a, which will give the 
swell of the column. 

To diminish a Column by Laths, upon the same Principle. 

In fig. D, the point e being found, as in fig. C, take and plow a rod d b, and lay the 
groove upon the axis of the column, and plow the describing rod upon the under side, 
and lay the groove upon a pin fixed at e, and fix a pin at g, to run in the groove upon 
the axis of the column, and the distance of the pencil at f equal to b a, then move the 
pencil a.tf it will describe the diminishing. 

How to describe the Column by another Method. 

Take the semi-diameter a b at bottom, and set the foot of your compass in the top 
at c, and cross the axis at 8, and draw the line a 8 on the outside, parallel to b 8 on the 
axis, and divide each of these lines into eight equal parts, and set the diameter a b at the 
bottom along the slant lines 1 1, 2 2, 3 3, &c, from the axis; this will also give the 
diminishing of the column. 

How to make a diminishing Rule. 

Divide the height of your rule into any number of equal parts, as 6; draw lines at 
right angles from these points across the rule, and divide the projection of the rule at 
the top ; that is, half of what the column diminishes ; into the same number of equal parts 
put a pin or brad-awl; lay a ruler from a to 5 ; mark the cross line at f; then lay a ruler 
from 4 to a, and mark the next cross line at e ; then lay the ruler from 3 to a, mark the 
next at d, and so on to the bottom ; bend a slip round these points, and draw the curve 
by it, will give a proper curve for the side of the column. 

Note. This is the readiest method, and gives the best curve of any that I have tried. 



/'///// 73 




Piatt 74. 




CYLINDRO-CYLINDRIC SASH-WORK, &c. 101 

PLATE 74. 

The Plan and Elevation of a circular Sash, in a circular Wall, being given; to find the Mould 
for the radial Bars, so that they shall be perpendicular to the Plan. 

Draw perpendiculars from the points 1, 1, 1, 1, &c, at A and B, in the radial bars, 
either equally divided, or taken at discretion down upon the plan, to 1, 2, 3, 4, 5, 6, 7, at 
C and D ; and draw a line from the first division upon the convex side parallel to the base ; 
then draw ordinates from 1, 1, 1, 1, &c, at right angles to the radial bars, at A and B, 
which being pricked from the plans at D and C, will give a mould for each bar; and 
the bevels upon the end will show the application of the moulds. 

To find the Veneer of the Arch-bar, or what is improperly called by Workmen Cot or Cod-bar. 

To avoid confusion, I have laid down the plan and elevation for the head of the sash 
under. The stretch-out of the veneer is got round, 1, 2, 3, 4, 5, 6, on the arch-bar, 
which being pricked from the small distance on the plan at M, will give the veneer above, 
at E. 

To find the Face-mould for the Sash-head. 

Divide the sash-head round, into any number of equal parts, at G, and draw them 
perpendicular to the base at H; draw the chord of one-half of the plan at H, and draw a 
line parallel to it to touch the plan upon the back side ; then the distance between these 
lines at H, will show what thickness of stuff the head is to be made out of; and from the 
intersecting points on the back side, draw perpendiculars from the base of the face mould, 
which being pricked from the elevation, as the figures direct, will give the face mould. 

To find the Moulds for giving the Form of the Head, perpendicular to the Plan. 

The base of L is got round the arch 1 2 3 4 5 6, at F, and the base of K is got round 
abed efg, also at F, and the heights of the ordinates of each are pricked either from H or 
/, which will give both moulds. 

By the same method, a circular achitrave, in a circular wall, may be got out of the 
solid. 

Note. The face mould at G must be applied in the same manner as in groins ; so that the sash-head must be 
bevelled by shifting the mould G, on each side, before you can apply the moulds JSTand L ; the black lines at AT and 
L are pricked from the plan, at //; these black lines will exactly coincide with the front of the rib when bent round ; 
a line being drawn by the other edge of the moulds, will be perpendicular over its plan, and the thickness of the 
sash-frame towards the inside will be found near enough by gauging from the outside. 



102 RAKING MOULDINGS, &c. 



PLATE 75. 

Construction of a circular headed architrave in a circular wall. See the description on 
the plate. 




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RAKING MOULDINGS, &c. 103 



PLATE 76. 

To describe the Angle Bars for Shop Fronts. 

In fig. A, B is a common bar, and C is the angle bar of the same thickness ; take 
the raking projection 1, 1, in C, and set the foot of your compass in 1 at B, and cross the 
middle of the base at the other 1; then draw the lines 2, 2; 3, 3, &c, parallel to 1, 1; 
then prick your bar at C from the ordinates so drawn at B, which being traced will give the 
angle bar. 

How to draw the Mitre Angle of a Commode Front for a Shop. 

In E divide the projection each way in a like number of equal parts, then the parallel 
lines continued each way will give the mitre. 

How to find the Raking Mouldings of a Pediment. 

In fig. F let the simarecta on the under side be the given moulding, and let lines be 
drawn upon the rake at discretion ; but if you please, let them be equally divided upon the 
simarecta, and drawn parallel to the rake ; then the mould at the middle being pricked off 
from these level lines at the bottom, will give the form of the face. The return moulding 
at the top must be pricked upon the rake, according to the letters. 

The cavetto, fig. C, is drawn in the same manner. 

N. B. If the middle moulding,^. F, be given, perpendiculars must be drawn to the 
top of it; then horizontal lines must be drawn over the mouldings at each end, with the 
same divisions as are over the mouldings ; and lines being drawn perpendicularly down, as 
above, will show how to trace the end mouldings. 



104 RAKING MOULDINGS, &c. 



PLATE 77. 

Raking Mouldings, and Mouldings of different Projections. 

Figures B and A show how to trace base mouldings for skirting to stairs, upon the 
same principles as shown in the last plate ; at the bottom are given two methods of mitring 
mouldings of different projections together. 



t'Uifr 7 




Ffatt 78 



This shews ///( ///i///ri/ of Ol/diyiii/j rort/nrs 

Fin- A. 




SIMILAR MOULDINGS, &c. 105 



PLATE 78. 

Given the Form of a Cm-nice, to draw it to a greater Proportion. 

In Jig. A, let the given height of the cornice be a d; set one foot of your compass in a, 
and cross the under side at b with that height, and from the point c draw the line c d at right 
angles to a b; then the height of all your mouldings will be the parts of a b, and the projec- 
tions the parts of c d in proportion. 

Note: af shows another height, c e its projection in proportion to that height. 

How to diminish a Cornice in the Proportion of a greater. 

Describe equilateral triangles on the base and projection as at D, and make if and i g 
equal to the intended height, and draw the Ivasfg across the triangle, which will give the 
heights in proportion to a b ; put the foot of your compass in b as a centre, and circle b c 
round b h, and draw the dotted line h i, cutting f g in k; then set off i e and i d, each equal to 
g k; draw d e; then take the divisions of e d, and set them from/" to m ; in the same order 
draw perpendiculars : it will give the diminished cornice at D. 

Another Method. 

At E, let the given height be a b, and draw the hypothenuse a g, and lines being 
squared up to a b, from the divisions of a g, will give the heights ; and if you draw the 
line g d at a right angle with a g, then d c will give the projection in proportion, when return- 
ing upon d c. 

Fig. C is the Method for hanging a Jib Boor. 

Let a c be the projection of the surbase or base moulding, and c the centre of the hinge; 
make a b equal to a c, and in the centre at c describe the arch b d e ; then the arch b d e will 
be the proper joint for the surbase to work in. 

The joint of the surbase or the base may also be straight, as you see by the dotted 
line touching the circle at the point b, as the tangent to it. 

14 



106 MOULDINGS UPON THE SPRING. 



PLATE 79. 

MOULDINGS UPON THE SPRING. 

To find the Sweep of a Moulding to be bent upon the Spring round a circular Cylinder. 

In fig. A, which stands upon a semicircular plan, make a c equal to the height of your 
moulding, and make a b equal to the projection ; describe the form of the moulding, and draw 
a dotted line to touch the face of it; then draw the line e d to meet in the centre of the body, 
at d, so as to keep your moulding to a sufficient parallel thickness ; from the centre d de- 
scribe the several concentric circles which are the arrises of the moulding required. 

How to find the Sweep of your Moulding when the Plan is a Segment. 
Complete the semicircle as in Plate I, fig. 1, then proceed as described in fig. A. 

Figures C and D show the method for bending a moulding round the inside, which is 
performed the same as above. 

The demonstration may easily be conceived from the covering of a cone. 



PI ah 79 



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SKY-LIGHTS. 107 



PLATE 80. 

To find the Length of the Hips of a Sky-light standing upon a square Plan, the Height being given. 

In fig. A, draw the diagonals a b, and c d ; they will bisect each other at right angles at e ; take a e 
for the base of any hip ; from e in e d make ef equal to the height of the sky-light ; join a,f and ajf will 
be the length of the hip required. 

To find the Backing of the Hip. 

Draw any line k i, at right angles to a e, the base of the hip rafter, cutting it in any point h, put the 
foot of your compass in h, as a centre, and with the other describe a circle to touch af, the hip rafter, to 
cut the base line a e, at g ; then draw g i and g k ; then the angle k g i will be the backing of the hip, as 
is shown by the bevel at B ; but the best way to work the hips is to apply a bevel to the parallel sides of 
the hips, as is shown at C, by making the other side of the bevel parallel to a e, the base of the hip. 

Note. The same lines will extend to any sky-light, whatever may be the form of its plan ; if it be any polygon, 
to find the length of the hip rafter, draw a line through any point in its base at right angles to it, so as to cut the 
two contiguous sides to that base, and on the said point as a centre describe a circle to touch the hip rafter from the 
points where this circle cuts the base line, draw two lines to meet the ends of the perpendicular line at the sides of 
the polygon; then the angle formed by these two lines will be the backing required: but perhaps a few more exam- 
ples will make it plainer than many words can. 

Fig. B is a sky-light, standing upon a rectangular base, having a ridge in the middle ; make c d upon 
the ridge line equal to half the width of a b ; then the angle b d a will be a right angle: every other re- 
quisite is the same as directed for fig. A. If these hips are to be mitred, the bevel at C shows the mitre. 

Fig. C is another sky-light, standing also upon a rectangular base ; but the hips all meet over the 
centre of the plan at e, and consequently the diagonals do not bisect each other at right angles; therefore 
take any base line, as a e, or e g, and make ef perpendicular to a e, from e, and equal to the height of the 
sky-light; and draw^a or fg, for the length of the hip, by drawing the line I m at right angles to a e. 
The backing will be found in the same manner as the others above. This sky-light will require two dif- 
ferent bevels D and E, to be applied to the parallel sides of the hip, which are both found from the back- 
ing by drawing the stocks of the bevel parallel to a e, the base of the hip. 

But if the hips are to be mitred together, F and G show the two bevels for the mitring each half, so 
that when put together shall form the proper backing. 

Fig. D is a sky-light standing upon an octangular plan, as is described in fig. 8, Plate 6, of the Geo- 
metry; the lengths of the hips and backing of the angle are found in the same manner as directed for others. 

Fig. £ is a sky-light whose plan is trapezoid ; upon each end as a diameter describe a semicircle to 
cut the ridge line; from these points draw lines to the extremities of their respective diameters, which will 
form a right angle for the base of the hips to stand upon ; the backing or mitring of the hips will be found 
as is described in fig. A and B. 



I'lfitt 81 



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CONCLUSION. 



In which are examined, by way of preventive Caution to the Student, several Methods, 
which are founded on wrong Principles, and better ones are here proposed. 



PLATE 81. 

Of the Ellipsis. 

THE old manner for intersecting all kinds of lines, applied to Gothic and elliptical 
figures, to this day is exceedingly useful in forming the ramps of stairs, or easing of any 
angle, as at G ; but when this is applied to elliptical figures, it is far from forming a true 
ellipsis, being too full at the ends, as at fig. D and E; and this is no certain rule for drawing 
an ellipsis, for the more divisions there are, the worse is the ellipsis ; as for example, fig. F 
is divided into double the number of parts as D ; it is plain that neither D nor C is an 
agreeable ellipsis, and C is much worse than D, which is contrary to general opinion; for I 
have been frequently told, the more parts it is divided into, the truer it is : but by this it ap- 
pears the more parts it is divided into, the worse it is : if this is doubted, try. Fig. A is an 
ellipsis drawn on true principles, as laid down in Plate 7, at C, of this book, and is here re- 
peated to be compared with the others. Figures B, C, and E, are ellipses drawn with a 
compass : I may call them representations, as it is impossible to draw them true with a com- 
pass ; there is no part of the curvature of an ellipsis that will exactly agree with any part of 
a circle, for in every quarter of an ellipsis, from the extremity of the transverse, the curva- 
ture in every succeeding part is continually flatter towards the extremity of the conjugate 
axis ; but yet there is a method to represent an ellipsis, which will differ very little from the 
truth, as is shown at figures B and E, which are both drawn by the same method ; fig. E 
or B is the nearest to the shape of fig. A ; fig. C, the method used by almost every author 
who has written upon the subject, as full at the ends, but not in so great a degree as fig. D 
and E. 



110 CONCLUSION. 



PLATE 82. 

Of Raking Mouldings. 

In Plate 76, fig. A, let the moulding at the bottom be given, and let the perpendicular 
height be divided into any number of equal parts at 6 ; likewise divide the perpendicular 
height of the top moulding into six ; and the face moulding into six, at right angles to the 
rake ; and let the ordinates of each be drawn through the equal divisions of each respective 
perpendicular, and pricked from the bottom, as the figures direct. It is evident that if the 
under moulding is composed of two quarters of a circle, the upper mouldings will be com- 
posed of two quarters of an ellipsis ; consequently, the return moulding at the top will be too 
quick upon the round, and likewise in the hollow. 

But if this demonstration should not be sufficient, let us suppose a plane parallel to the 
arrises of the moulding, and perpendicular to the plane of projection, to pass through the 
point 1, this plane will be represented by a straight line; therefore let a dotted line be con- 
tinued from l,in the given moulding, parallel to the rake; then it will be evidently seen, that 
this dotted line corresponds with neither the face nor the return moulding ; for in the face 
moulding it falls between the points 1 and 2 ; and in the top moulding it falls almost at the 
point 2 ; whereas it should only come to the point 1 in each ; but the horizontal projection 
from 2 to 2, at the top, ought to be equal to 1 1 at the bottom ; but it is much greater; there- 
fore this method is false, and they will not mitre together. 

I shall also notice another method used by some authors, see figure B, at D and E, where 
they are pricked perpendicular to their chords, in the middle, which is also false ; but if they 
are pricked as at B and C, on the rake, will be exceedingly near, if described with a compass 
through three points. 



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PLATE 83. 

Of diminishing of Columns. 

The method for drawing a column, described by some authors, and which is properly 
called a conchoid column, is not only very inconvenient on account of the cumbersome instru- 
ment which is necessary to find the curve for practice, but also the appearance is, I think, 
less graceful than when produced by other methods. The conchoid curve, or column, is 
concave towards the bottom, and convex towards the top ; and if this curve was infinitely 
extended, it would never meet the axis ; which shows it to be different from the elliptic 
curve or column, as some have called it. 

The column called hyperbolic, Plate 18, is not so named from the general properties of 
the conic hyperbola (because there may be an infinite number of hyperbolas standing upon 
the same base, having one common vertex, which will all be contained between a triangle 
and a parabola, according as its axis is longer or shorter), but because it will nearly coin- 
cide with some of these hyperbolas. This curve has been known among workmen, and 
by them has been mistaken for an elliptic curve; to refute which I have, on the same 
plate, shown a true elliptic column for comparison ; the lines of their curvature are continued 
only to show their true figure; either of these is a more commodious method than the con- 
choid. 

The method which I recommend as easiest in practice for diminishing of columns is 
already described on Plate 73, by means of a diminishing rule, which is infinitely more con- 
venient than the trammel, and which to my eye, also produces a pleasant contour : but as 
this will depend on the fancy of the architect, the workman will find some of the methods 
shown will answer his purpose for any curve. The conchoid is flattest at the top, the hyper- 
bolic is a little quicker, the parabolic is still more so, and the elliptic is the most quick. 



INDEX. 



ACUTE angle, defined 10 

Acute angled triangle, defined 10 

Altitude of a figure, defined - - - 11 

Angle, acute, defined 10 

Angle bracket ----- 42 

Angle, definition 10 

Angle, how bisected - - - - 12 

Angle, how denoted - - - - H 

Angle, how made equal to a given angle - 12 

Angle, how measured - - - - 11 

Angle, obtuse, defined 10 

Angle bars for shop fronts ... 103 

Angle ribs 32 

Angle, right, defined 10 

Angle mould of a groin 29 

Angles, kinds of ----- 10 

Arc of a circle, defined - - - - 11 

Arc of a circle, how its length may be found 13 

Arch bar 101 

Arches, how described 28 

Ascending groins 30 

15 



B 

Bevelling the edges of ribs 28 

Boarding of polygonal figures 48 
Boarding of domes, when the boards are bent 

vertically ----- 48 
Boarding of domes, when the boards are bent 

horizontally - - - - - . 49 

Boarding of an ellipsoidal dome - - 50 

Body, or body range of a groin - - 29 

Butt joints in hand-railing, how made - 83 

C 

Centre of a circle, defined - - - 11 

Centring to groins ----- 29 

Chord in a circle, defined - - - 11 

Circle, defined ----- 11 

Circumference, defined - - - - 11 
Circumference of a circle, how described 

through three given points - - 13 

Column*, how diminished ... 100 

Common bracket ----- 42 

Commode front 103 



114 



INDEX. 







PAGE 


Cone, defined ... 


- 


17 


Conic sections - 


- 


17 


Cornices, proportioned 


- 


- 105 


Cot, or cod-bar - 


- 


- 101 


Cove bracket - - - - 


- 


42 


Covering spherical domes 


- 


48 


Coves ----- 


- 


42 


Cradling to groins - - - 


- 


- 30, 38 


Cradling for the heads of niches 


- 


39 


Cradling pendentives 


- 


43 


Curb, how formed in the ceilings 


of chi 


irches 28 


Curtail step, and riser 


- 


90 


Curve line, defined - - - 


- 


10 


Cylindric sections - - - 


- 


19 


Cylindric groins 


- 


30 


Cylindro-cylindric arch, defined 


- 


35 


Cylindro- cylindric sash-work - 


- 


- 101 



D 

Decagon, defined - - - - - 11 

Designs for roofs ----- 47 

Diameter of a circle, defined - - - 11 

Diminishing rule ----- 100 

Diminution of columns - - - 100, 111 

Dodecagon, defined 11 

E 

Ellipsis, how circumscribed about a rectangle 
with its axis parallel to, and in the same 

ratio, as the sides of the rectangle - 16 

Ellipsis, a conic section 17 

Ellipsis, how described from the cone - 17 



18 



16 



Ellipsis, how described by the intersection of 

straight lines 

Ellipsis, the two axes of an, being given, to 
describe the curve with compasses, by 
ordinates, by a string, and by a trammel 
Ellipsis, being given, to find its axes and centre 16 
Ellipsis, the length of an, being given, to find 
the other axis, so that the two shall be in 
a given ratio ----- 16 
Ellipsis, observations thereon - - - 109 
Ellipsoidal-headed niche 42 

Equilateral triangle, defined 10 

Equilateral triangle, a regular figure - - 11 
Equilateral rectangle, defined 10 

Equilateral parallelogram, defined - - 11 
Equilateral triangle, how constructed upon 

a given line ----- 13 



Face moulds - - 81, 83, 86, 89, 94, 97 

Falling moulds - - 80, 85, 88, 97, 98 

Four sided figures, how called - - - 11 



G 



Geometry 
Globe, its 
Groins, how formed 



Globe, its sections 



H 



Hand-railing - - - 
Height of a figure, defined 
Hip roof, how constructed 



9 to 22 

21 

29 to 34 



75 to 90, 98, 99 



11 
45 



INDEX. 



115 



PAGE 

Hip sky-lights 105 

Hyberbola constructed from the cone - 17 

I 

Intersection of the angles of a groin 30 

Irregular polygon, how constructed - - 14 

Isosceles triangle 10 



Jack rafter, how fitted upon the hip - 45 

Jack ribs of a groin .... 29, 30 

Jib door, how hung .... 105 

K 

Kirb lights for church work 28 



Level landing .... 93, 102 

Line, bisected ----- 12 

Line, defined ------ 10 

Line divided as another 16 

Lines, the several kinds 10 

Lining of a soffit 23 

M 

Mean proportion, how found between two 

straight lines ----- 13 

Mitre bracket of a cornice 42 

Mouldings upon the spring ... 106 



N 



Niches 



- 39, 42 



O 

Oblique lines 10 

Obtuse angle, defined 10 

Obtuse angled triangle, defined 10 

Octagon, defined - - - - - 11 

Octagon made out of a given square - 16 



Parabola, a conic section 17 

Parabola, how described from a cone - 17 

Parallelogram, defined - - - - 11 

Parallel, defined 10 

Pediment, raking mouldings of - - - 102 

Pendentives ------ 43 

Pentagon, defined - - - - - 11 

Pentagon,. how made 13 

Perpendicular, defined 10 

Perpendicular, to draw a 12 

Perpendicular, to let fall a 12 
Perpendicular at the end of a right line, to 

draw a 12 

Plane, defined 10 

Plane figures, how bounded 10 

Plane figures, how named 10 

Plane of the segment of a cylinder, defined 19 

Point, definition of a - - - - 10 

Polygon, defined - - - - - H 

Polygon, irregular, defined - - - 11 
Polygon, how constructed upon a given right 

line 13 

Polygon, irregular, being given to construct, 

and then equal and similar 14 



116 



INDEX. 



Polygon roofs 48 

Purline, how fitted to the hip rafter - - 45 

Q 

Quadrangle, defined -.._.; . jo 

Quadrangle, how made equal to a given one 14 

Quadrant of a circle, defined - - - 11 

Quadrilateral, defined 10 

R 

Raking mouldings - 102,110 

Radius of a circle, defined - - • 11 
Ranging the edges of ribs standing upon 

curbs in churches 28 
Ranging the ribs of groins ... 32, 33 

Rectangle, defined 11 

Rectangle, equilateral, defined - - - 11 
Rectangle, how made equal to a given tri- 
angle ------ 14 

Rhomboid, defined - - - - - 11 

Rhombus, defined ----- 11 

Rib of the front of the head of a niche - 39 
Ribs of the back of the head of a niche - 39 
Ribs of the spheric headed niches, how de- 
scribed ------ 40 

Ribs of spheric headed niches, how placed 

upon the front rib - - - - 40 

Ribs of spheric headed niches, how bevelled 

upon the front rib - - - - 40 

Right angle, defined 10 

Right angled triangle, defined 10 

Regular pentagon, defined - - - 11 



100, 111 


46 


45 to 50 


64 to 74 



Rule for diminishing columns 
Roof in ledgement - 
Roofing - - - - 
Roofs, designs for - 



S 
Sash head, circular in circular wall - - 101 
Scalene triangle, defined 10 

Scroll of a hand-rail, how drawn - - 87 
Seat of the quoins of a groin 29 

Section of a hemisphere cut by a cylinder 
surface at right angles to the plane of 
the segment ----- 21 
Section of a cylinder, defined 19 

Section of a sphere, what 21 

Section of any body generated by rotation 

round a fixed axis 21 

Section of a hemisphere cut by a plane at right 

angles to the plane of the segment - 21 
Sections of a cylinder 19 

Sections of a sphere - - - - 21 

Sector of a circle, defined - - - 11 

Segment of a circle, defined - - - 11 

Segment of a circle, how constructed to 
any chord, and versed sine by a com- 
pass, or by the intersection of straight 

lines 15, 18 

Segment of a cylinder being given, to cut 
it by a plane through a given line on 
the plane of a segment, and to make 
any given angle with that plane - 20 

Semicircle, defined - - - - - 11 



INDEX. 



117 





PAGE 


Sky-lights - 


- 107 


Soffits, linings of 


23 to 27 


Solid, definition of a 


10 


Spherical domes - 


29 to 50 


Spheric sections - 


21 


Spheric niches - 


40 


Springing of a pendentive 


43 


Spheroidal domes - 


50 


Spring of the plank - 


- 82, 92 


Square, defined - 


11 


Square, how made upon a given right line- 13 


Square, how made equal to a given 


rect- 


angle - 


14 



Square, how made equal to two given 

squares ------ 14 

Stair-casing 91 to 97 

Straight line, defined 10 

Stretch-out line ----- 23 
Strength of timber - - - - 51 to 63 
Superfices, kinds of 10 

Superfices, definition of - - - - 10 
Superfices, plane, definition 10 



Superfices, curved 



Tangent, defined 

Tangent to a circle, how drawn 

Tangent to a circle being given, to find 

the. point of contact - - - - 
Theoretical part of Geometry, how founded 
Trapezium, defined - 

Trapezoid, defined - - - - - 
Triangle, equilateral, defined - - - 
Triangle, isosceles, defined - 
Triangle, scalene, defined - 

Triangle, right angled, definition of a 
Triangle, oblique, definition of a 
Triangle, acute angled, definition of a 
Triangle, obtuse angled, definition of a 
Triangle, how constructed of three given 

right lines - - - - - 



U 



Undecagon, defined 



10 



9 
13 

13 
9 
11 
11 
10 
10 
10 
10 
10 
10 
10 

14 



11 



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